NCERT Solutions: Class 12 Maths Chapter 7 - Integrals

Since $\dfrac{d}{dx}(\cos 2x)=-2\sin 2x$, an anti derivative of $\sin 2x$ is $-\dfrac12\cos 2x+C$.

Since $\dfrac{d}{dx}(\sin 3x)=3\cos 3x$, an anti derivative of $\cos 3x$ is $\dfrac13\sin 3x+C$.

Since $\dfrac{d}{dx}(e^{2x})=2e^{2x}$, an anti derivative of $e^{2x}$ is $\dfrac12e^{2x}+C$.

Since $\dfrac{d}{dx}(ax+b)^3=3a(ax+b)^2$, an anti derivative of $(ax+b)^2$ is $\dfrac{(ax+b)^3}{3a}+C$.

Integrate term by term: $\int \sin 2x\,dx=-\dfrac12\cos 2x$ and $\int -4e^{3x}\,dx=-\dfrac43e^{3x}$. Thus the anti derivative is $-\dfrac12\cos 2x-\dfrac43e^{3x}+C$.

$\int (4e^{3x}+1)\,dx=4\int e^{3x}\,dx+\int 1\,dx=\dfrac43e^{3x}+x+C$.

Simplify the integrand: $x^2\left(1-\dfrac1{x^2}\right)=x^2-1$. Hence $\int (x^2-1)\,dx=\dfrac{x^3}{3}-x+C$.

Integrating term by term, $\int (ax^2+bx+c)\,dx=a\dfrac{x^3}{3}+b\dfrac{x^2}{2}+cx+C$.

$\int (2x^2+e^x)\,dx=2\dfrac{x^3}{3}+e^x+C=\dfrac{2x^3}{3}+e^x+C$.

Expand: $\left(\sqrt{x}-\dfrac1{\sqrt{x}}\right)^2=x-2+\dfrac1x$. Therefore $\int \left(x-2+\dfrac1x\right)dx=\dfrac{x^2}{2}-2x+\log|x|+C$.

Divide each term by $x^2$: $\dfrac{x^3+5x^2-4}{x^2}=x+5-4x^{-2}$. Hence the integral is $\dfrac{x^2}{2}+5x+4x^{-1}+C=\dfrac{x^2}{2}+5x+\dfrac4x+C$.

Write the integrand as $x^{5/2}+3x^{1/2}+4x^{-1/2}$. Therefore the integral is $\dfrac{2}{7}x^{7/2}+2x^{3/2}+8x^{1/2}+C$.

Factor the numerator: $x^3-x^2+x-1=(x-1)(x^2+1)$. Hence the integrand is $x^2+1$, and the integral is $\dfrac{x^3}{3}+x+C$.

$(1-x)\sqrt{x}=x^{1/2}-x^{3/2}$. Thus $\int (1-x)\sqrt{x}\,dx=\dfrac{2}{3}x^{3/2}-\dfrac{2}{5}x^{5/2}+C$.

Expand the integrand: $\sqrt{x}(3x^2+2x+3)=3x^{5/2}+2x^{3/2}+3x^{1/2}$. Hence the integral is $\dfrac{6}{7}x^{7/2}+\dfrac{4}{5}x^{5/2}+2x^{3/2}+C$.

Integrate term by term: $\int 2x\,dx=x^2$, $\int -3\cos x\,dx=-3\sin x$, and $\int e^x\,dx=e^x$. Thus the integral is $x^2-3\sin x+e^x+C$.

Termwise integration gives $\int 2x^2\,dx=\dfrac{2x^3}{3}$, $\int -3\sin x\,dx=3\cos x$, and $\int 5\sqrt{x}\,dx=\dfrac{10}{3}x^{3/2}$. Hence the result.

$\sec x(\sec x+\tan x)=\sec^2x+\sec x\tan x$. Since $\int \sec^2x\,dx=\tan x$ and $\int \sec x\tan x\,dx=\sec x$, the integral is $\tan x+\sec x+C$.

Since $\dfrac{\sec^2x}{\cosec^2x}=\dfrac{1/\cos^2x}{1/\sin^2x}=\tan^2x=\sec^2x-1$, the integral is $\tan x-x+C$.

$\dfrac{2-3\sin x}{\cos^2x}=2\sec^2x-3\tan x\sec x$. Therefore the integral is $2\tan x-3\sec x+C$.

1. i. $\dfrac13x^{1/3}+2x^{1/2}+C$
2. ii. $\dfrac23x^{2/3}+\dfrac12x^2+C$
3. iii. $\dfrac23x^{3/2}+2x^{1/2}+C$
4. iv. $\dfrac32x^{3/2}+\dfrac12x^2+C$

$\int \left(\sqrt{x}+\dfrac1{\sqrt{x}}\right)dx=\int (x^{1/2}+x^{-1/2})dx=\dfrac23x^{3/2}+2x^{1/2}+C$. Hence the correct option is (iii).

1. i. $x^4+\dfrac1{x^3}-\dfrac{129}{8}$
2. ii. $x^3+\dfrac1{x^4}+\dfrac{129}{8}$
3. iii. $x^4+\dfrac1{x^3}+\dfrac{129}{8}$
4. iv. $x^3+\dfrac1{x^4}-\dfrac{129}{8}$

Integrating, $f(x)=\int\left(4x^3-\dfrac3{x^4}\right)dx=x^4+\dfrac1{x^3}+C$. Since $f(2)=0$, $16+\dfrac18+C=0$, so $C=-\dfrac{129}{8}$. Hence $f(x)=x^4+\dfrac1{x^3}-\dfrac{129}{8}$, option (i).

Put $u=1+x^2$. Then $du=2x\,dx$. Hence $\int \dfrac{2x}{1+x^2}\,dx=\int \dfrac{du}{u}=\log|u|+C=\log(1+x^2)+C$.

Put $u=\log x$. Then $du=dx/x$. Therefore $\int \dfrac{(\log x)^2}{x}\,dx=\int u^2\,du=\dfrac{u^3}{3}+C=\dfrac{(\log x)^3}{3}+C$.

Since $x+x\log x=x(1+\log x)$, put $u=1+\log x$. Then $du=dx/x$. Thus the integral is $\int \dfrac{du}{u}=\log|u|+C=\log|1+\log x|+C$.

Put $u=\cos x$. Then $du=-\sin x\,dx$. Hence $\int \sin x\sin(\cos x)\,dx=-\int \sin u\,du=\cos u+C=\cos(\cos x)+C$.

Put $u=\sin(ax+b)$. Then $du=a\cos(ax+b)\,dx$. Thus $\int \sin(ax+b)\cos(ax+b)\,dx=\dfrac1a\int u\,du=\dfrac{\sin^2(ax+b)}{2a}+C$.

Put $u=ax+b$. Then $du=a\,dx$. Therefore $\int \sqrt{ax+b}\,dx=\dfrac1a\int u^{1/2}\,du=\dfrac{2}{3a}(ax+b)^{3/2}+C$.

Put $u=x+2$, so $x=u-2$ and $dx=du$. Then $\int x\sqrt{x+2}\,dx=\int (u-2)u^{1/2}\,du=\dfrac25u^{5/2}-\dfrac43u^{3/2}+C=\dfrac{2}{15}(3x-4)(x+2)^{3/2}+C$.

Put $u=1+2x^2$. Then $du=4x\,dx$. Hence $\int x\sqrt{1+2x^2}\,dx=\dfrac14\int u^{1/2}\,du=\dfrac16(1+2x^2)^{3/2}+C$.

Put $u=x^2+x+1$. Then $du=(2x+1)dx$ and $(4x+2)dx=2du$. Thus the integral is $2\int u^{1/2}du=\dfrac43u^{3/2}+C=\dfrac43(x^2+x+1)^{3/2}+C$.

Put $t=\sqrt{x}$, so $x=t^2$ and $dx=2t\,dt$. Then $x-\sqrt{x}=t(t-1)$ and the integral becomes $\int \dfrac{2t}{t(t-1)}dt=2\int \dfrac{dt}{t-1}=2\log|t-1|+C=2\log|\sqrt{x}-1|+C$.

Put $u=x+4$, so $x=u-4$. Then $\int \dfrac{x}{\sqrt{x+4}}dx=\int (u^{1/2}-4u^{-1/2})du=\dfrac23u^{3/2}-8u^{1/2}+C=\dfrac23(x-8)\sqrt{x+4}+C$.

Put $u=x^3-1$. Then $du=3x^2dx$ and $x^5dx=x^3x^2dx=(u+1)du/3$. Hence the integral is $\dfrac13\int (u^{4/3}+u^{1/3})du=\dfrac{u^{7/3}}{7}+\dfrac{u^{4/3}}{4}+C$.

Put $u=2+3x^3$. Then $du=9x^2dx$. Therefore $\int \dfrac{x^2}{(2+3x^3)^3}dx=\dfrac19\int u^{-3}du=-\dfrac{1}{18u^2}+C=-\dfrac{1}{18(2+3x^3)^2}+C$.

Put $u=\log x$. Then $du=dx/x$. Hence $\int \dfrac{dx}{x(\log x)^m}=\int u^{-m}du=\dfrac{u^{1-m}}{1-m}+C=\dfrac{(\log x)^{1-m}}{1-m}+C$.

Put $u=9-4x^2$. Then $du=-8x\,dx$. Thus $\int \dfrac{x}{9-4x^2}dx=-\dfrac18\int \dfrac{du}{u}=-\dfrac18\log|9-4x^2|+C$.

Put $u=2x+3$. Then $du=2dx$. Hence $\int e^{2x+3}dx=\dfrac12\int e^u du=\dfrac12e^{2x+3}+C$.

$\dfrac{x}{e^{x^2}}=xe^{-x^2}$. Put $u=-x^2$, so $du=-2x\,dx$. Then $\int xe^{-x^2}dx=-\dfrac12e^{-x^2}+C$.

Put $u=\tan^{-1}x$. Then $du=dx/(1+x^2)$. Thus $\int \dfrac{e^{\tan^{-1}x}}{1+x^2}dx=\int e^u du=e^u+C=e^{\tan^{-1}x}+C$.

Since $\dfrac{d}{dx}\log(e^x+e^{-x})=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}=\dfrac{e^{2x}-1}{e^{2x}+1}$, the required integral is $\log(e^x+e^{-x})+C$.

The derivative of $e^{2x}+e^{-2x}$ is $2e^{2x}-2e^{-2x}=2(e^{2x}-e^{-2x})$. Hence the integral is $\dfrac12\log(e^{2x}+e^{-2x})+C$.

Use $\tan^2u=\sec^2u-1$. Then $\int \tan^2(2x-3)dx=\int(\sec^2(2x-3)-1)dx=\dfrac12\tan(2x-3)-x+C$.

Put $u=7-4x$. Then $du=-4dx$. Hence $\int \sec^2(7-4x)dx=-\dfrac14\int \sec^2u\,du=-\dfrac14\tan(7-4x)+C$.

Put $u=\sin^{-1}x$. Then $du=dx/\sqrt{1-x^2}$. Therefore the integral is $\int u\,du=\dfrac12u^2+C=\dfrac12(\sin^{-1}x)^2+C$.

The derivative of $6\cos x+4\sin x$ is $-6\sin x+4\cos x=2(2\cos x-3\sin x)$. Hence the integral is $\dfrac12\log|6\cos x+4\sin x|+C$.

Since $\sec^2x=1/\cos^2x$, put $u=1-\tan x$. Then $du=-\sec^2x\,dx$. The integral becomes $-\int u^{-2}du=\dfrac1u+C=\dfrac{1}{1-\tan x}+C$.

Put $t=\sqrt{x}$. Then $dx=2t\,dt$. Hence $\int \dfrac{\cos\sqrt{x}}{\sqrt{x}}dx=\int \dfrac{\cos t}{t}2t\,dt=2\sin t+C=2\sin\sqrt{x}+C$.

Put $u=\sin 2x$. Then $du=2\cos 2x\,dx$. Therefore $\int \sqrt{\sin 2x}\cos 2x\,dx=\dfrac12\int u^{1/2}du=\dfrac13u^{3/2}+C$.

Put $u=1+\sin x$. Then $du=\cos x\,dx$. Hence the integral is $\int u^{-1/2}du=2\sqrt{u}+C=2\sqrt{1+\sin x}+C$.

Put $u=\log\sin x$. Then $du=\cot x\,dx$. Therefore $\int \cot x\log\sin x\,dx=\int u\,du=\dfrac12u^2+C=\dfrac12(\log\sin x)^2+C$.

Put $u=1+\cos x$. Then $du=-\sin x\,dx$. Hence the integral is $-\int \dfrac{du}{u}=-\log|1+\cos x|+C$.

Put $u=1+\cos x$. Then $du=-\sin x\,dx$. Thus the integral is $-\int u^{-2}du=\dfrac1u+C=\dfrac{1}{1+\cos x}+C$.

$\dfrac1{1+\cot x}=\dfrac{\sin x}{\sin x+\cos x}=\dfrac12+\dfrac12\dfrac{\sin x-\cos x}{\sin x+\cos x}$. Since $d(\sin x+\cos x)=(\cos x-\sin x)dx$, the integral is $\dfrac{x}{2}-\dfrac12\log|\sin x+\cos x|+C$.

$\dfrac1{1-\tan x}=\dfrac{\cos x}{\cos x-\sin x}=\dfrac12+\dfrac12\dfrac{\cos x+\sin x}{\cos x-\sin x}$. Since $d(\cos x-\sin x)=-(\sin x+\cos x)dx$, the integral is $\dfrac{x}{2}-\dfrac12\log|\cos x-\sin x|+C$.

Put $u=\tan x$. Since $du=\sec^2x\,dx=dx/\cos^2x$ and $\sin x\cos x=u\cos^2x$, the integrand becomes $u^{-1/2}du$. Thus the integral is $2\sqrt{u}+C=2\sqrt{\tan x}+C$.

Put $u=1+\log x$. Then $du=dx/x$. Hence the integral is $\int u^2du=\dfrac{u^3}{3}+C=\dfrac{(1+\log x)^3}{3}+C$.

Put $u=x+\log x$. Then $du=(1+1/x)dx=\dfrac{x+1}{x}dx$. Therefore the integral is $\int u^2du=\dfrac{(x+\log x)^3}{3}+C$.

Put $u=\tan^{-1}x^4$. Then $du=\dfrac{4x^3}{1+x^8}dx$. Hence the integral is $\dfrac14\int \sin u\,du=-\dfrac14\cos u+C=-\dfrac14\cos(\tan^{-1}x^4)+C$.

1. i. $10^x-x^{10}+C$
2. ii. $10^x+x^{10}+C$
3. iii. $(10^x-x^{10})^{-1}+C$
4. iv. $\log(10^x+x^{10})+C$

The derivative of $x^{10}+10^x$ is $10x^9+10^x\log_e 10$. Hence the integral is $\log(x^{10}+10^x)+C$, which is option (iv).

1. i. $\tan x+\cot x+C$
2. ii. $\tan x-\cot x+C$
3. iii. $\tan x\cot x+C$
4. iv. $\tan x-\cot 2x+C$

Since $\dfrac{1}{\sin^2x\cos^2x}=\sec^2x+\cosec^2x$, the integral is $\tan x-\cot x+C$. Hence the correct option is (ii).

Use $\sin^2u=\dfrac{1-\os 2u}{2}$. Then $\int \sin^2(2x+5)dx=\dfrac{x}{2}-\dfrac12\int \cos(4x+10)dx=\dfrac{x}{2}-\dfrac{\sin(4x+10)}{8}+C$.

Using $\sin A\cos B=\dfrac12[\sin(A+B)+\sin(A-B)]$, the integrand is $\dfrac12(\sin 7x-\sin x)$. Hence the integral is $-\dfrac1{14}\cos 7x+\dfrac12\cos x+C$.

Use product-to-sum twice: $\cos2x\cos6x=\dfrac12(\cos8x+\cos4x)$. Multiplying by $\cos4x$ gives $\dfrac14(\cos12x+\cos4x)+\dfrac14(1+\os8x)$. Integrating term by term gives the stated result.

Put $u=2x+1$, so $dx=du/2$. Then $\int\sin^3(2x+1)dx=\dfrac12\int \sin u(1-\cos^2u)du=-\dfrac12\cos u+\dfrac16\cos^3u+C$.

Since $\sin^3x\cos^3x=\dfrac18\sin^3 2x$, put $u=2x$. Then the integral is $\dfrac1{16}\int \sin^3u\,du=-\dfrac{\cos 2x}{16}+\dfrac{\cos^3 2x}{48}+C$.

Using product-to-sum, $\sin x\sin3x=\dfrac12(\cos2x-\cos4x)$. Hence the integrand is $\dfrac14\sin4x-\dfrac14\sin6x+\dfrac14\sin2x$. Integrating gives $-\dfrac{\cos4x}{16}+\dfrac{\cos6x}{24}-\dfrac{\cos2x}{8}+C$.

$\sin4x\sin8x=\dfrac12(\cos4x-\cos12x)$. Therefore the integral is $\dfrac12\left(\dfrac{\sin4x}{4}-\dfrac{\sin12x}{12}\right)+C=\dfrac{\sin4x}{8}-\dfrac{\sin12x}{24}+C$.

$\dfrac{1-\cos x}{1+\cos x}=\tan^2(x/2)$. Put $u=x/2$, so $dx=2du$. Then the integral is $2\int\tan^2u\,du=2(\tan u-u)+C=2\tan\dfrac{x}{2}-x+C$.

$\dfrac{\cos x}{1+\os x}=1-\dfrac1{1+\os x}=1-\dfrac12\sec^2(x/2)$. Hence the integral is $x-\tan\dfrac{x}{2}+C$.

Use $\sin^4x=\left(\dfrac{1-\cos2x}{2}\right)^2=\dfrac{3-4\cos2x+\os4x}{8}$. Integrating gives $\dfrac{3x}{8}-\dfrac{\sin2x}{4}+\dfrac{\sin4x}{32}+C$.

Since $\cos^4u=\dfrac{3+4\cos2u+\os4u}{8}$, with $u=2x$ we get $\cos^4 2x=\dfrac{3+4\cos4x+\os8x}{8}$. Integrating term by term gives the answer.

$\dfrac{\sin^2x}{1+\os x}=\dfrac{(1-\os x)(1+\os x)}{1+\os x}=1-\os x$. Thus the integral is $x-\sin x+C$.

Use $\cos2x-\cos2\alpha=2(\cos^2x-\cos^2\alpha)=2(\cos x-\os\alpha)(\cos x+\os\alpha)$. The integrand becomes $2(\cos x+\os\alpha)$, whose integral is $2\sin x+2x\cos\alpha+C$.

Since $1+\in2x=(\sin x+\os x)^2$, put $u=\sin x+\os x$. Then $du=(\cos x-\in x)dx$. Thus the integral is $\int u^{-2}du=-\dfrac1u+C=-\dfrac{1}{\sin x+\os x}+C$.

Write $\tan^3 2x\sec2x=(\sec^2 2x-1)\tan2x\sec2x$. Put $u=\sec2x$, so $du=2\sec2x\tan2x\,dx$. The integral becomes $\dfrac12\int(u^2-1)du=\dfrac{u^3}{6}-\dfrac{u}{2}+C$.

Since $\int \tan u\,du=\log|\sec u|+C$, putting $u=4x$ gives $\int \tan4x\,dx=\dfrac14\log|\sec4x|+C$.

Split the fraction: $\dfrac{\sin^3x}{\sin^2x\cos^2x}+\dfrac{\cos^3x}{\sin^2x\cos^2x}=\dfrac{\sin x}{\cos^2x}+\dfrac{\cos x}{\sin^2x}$. These integrate to $\sec x$ and $-\cosec x$, respectively.

$\dfrac{\cos2x}{2\sin^2x\cos^2x}=\dfrac{\cos^2x-\sin^2x}{2\sin^2x\cos^2x}=\dfrac12(\cosec^2x-\sec^2x)$. Hence the integral is $-\dfrac12\cot x-\dfrac12\tan x+C$.

Put $t=\tan x$. Then $dx=dt/(1+t^2)$, $\sin x=t/\sqrt{1+t^2}$ and $\cos x=1/\sqrt{1+t^2}$. The integrand becomes $(t^{-3}+t^{-1})dt$. Thus the integral is $-\dfrac1{2t^2}+\log|t|+C=\log|\tan x|-\dfrac{1}{2\tan^2x}+C$.

Since $\cos2x=(\cos x-\in x)(\cos x+\in x)$, the integrand is $\dfrac{\cos x-\in x}{\cos x+\in x}$. Put $u=\cos x+\in x$; then $du=(\cos x-\in x)dx$. Hence the integral is $\log|u|+C$.

Using the principal-value identity $\sin^{-1}(\cos x)=\dfrac{\pi}{2}-x$ on the usual interval, the integral is $\int\left(\dfrac{\pi}{2}-x\right)dx=\dfrac{\pi x}{2}-\dfrac{x^2}{2}+C$.

Since $\tan(x-a)-\tan(x-b)=\dfrac{\sin(b-a)}{\cos(x-a)\cos(x-b)}$, the integrand equals $\dfrac{\tan(x-a)-\tan(x-b)}{\sin(b-a)}$. Integrating gives $\dfrac{1}{\sin(b-a)}[-\log|\cos(x-a)|+\log|\cos(x-b)|]+C$.

1. i. $\tan x+\cot x+C$
2. ii. $\tan x+\cosec x+C$
3. iii. $-\tan x+\cot x+C$
4. iv. $\tan x+\sec x+C$

The integrand is $\sec^2x-\cosec^2x$. Therefore the integral is $\tan x+\cot x+C$, so option (i) is correct.

1. i. $-\cot(e^x x)+C$
2. ii. $\tan(xe^x)+C$
3. iii. $\tan(e^x)+C$
4. iv. $\cot(e^x)+C$

Put $u=xe^x$. Then $du=e^x(1+x)dx$. The integral becomes $\int \sec^2u\,du=\tan u+C=\tan(xe^x)+C$. Hence option (ii) is correct.

Put $u=x^3$. Then $du=3x^2dx$. Hence $\int \dfrac{3x^2}{x^6+1}dx=\int \dfrac{du}{u^2+1}=\tan^{-1}u+C=\tan^{-1}(x^3)+C$.

Put $u=2x$, so $dx=du/2$. The integral becomes $\dfrac12\int \dfrac{du}{\sqrt{1+u^2}}=\dfrac12\log|u+\sqrt{1+u^2}|+C$.

Put $u=2-x$, so $du=-dx$. Then the integral is $-\int \dfrac{du}{\sqrt{u^2+1}}=-\log|u+\sqrt{u^2+1}|+C$.

Put $u=5x$, so $dx=du/5$. Then $\int \dfrac{dx}{\sqrt{9-25x^2}}=\dfrac15\int \dfrac{du}{\sqrt{3^2-u^2}}=\dfrac15\sin^{-1}\left(\dfrac{5x}{3}\right)+C$.

Put $u=x^2$. Then $du=2x\,dx$. The integral is $\dfrac32\int \dfrac{du}{1+2u^2}=\dfrac{3}{2\sqrt2}\tan^{-1}(\sqrt2u)+C$.

Put $u=x^3$, so $du=3x^2dx$. Then the integral is $\dfrac13\int \dfrac{du}{1-u^2}=\dfrac16\log\left|\dfrac{1+u}{1-u}\right|+C$.

Split the integral as $\int \dfrac{x}{\sqrt{x^2-1}}dx-\int \dfrac{dx}{\sqrt{x^2-1}}$. These are $\sqrt{x^2-1}$ and $\log|x+\sqrt{x^2-1}|$, respectively, giving the result.

Put $u=x^3$, so $du=3x^2dx$. Then $\int \dfrac{x^2}{\sqrt{x^6+a^6}}dx=\dfrac13\int \dfrac{du}{\sqrt{u^2+a^6}}=\dfrac13\log|u+\sqrt{u^2+a^6}|+C$.

Put $u=\tan x$. Then $du=\sec^2x\,dx$. The integral becomes $\int \dfrac{du}{\sqrt{u^2+4}}=\log|u+\sqrt{u^2+4}|+C$.

Complete the square: $x^2+2x+2=(x+1)^2+1$. Hence the integral is $\log|x+1+\sqrt{(x+1)^2+1}|+C$.

Since $9x^2+6x+5=(3x+1)^2+4$, put $u=3x+1$. Then $dx=du/3$, and the integral becomes $\dfrac13\int \dfrac{du}{u^2+4}=\dfrac16\tan^{-1}(u/2)+C$.

Complete the square: $7-6x-x^2=16-(x+3)^2$. Therefore $\int \dfrac{dx}{\sqrt{7-6x-x^2}}=\sin^{-1}\left(\dfrac{x+3}{4}\right)+C$.

$(x-1)(x-2)=\left(x-\dfrac32\right)^2-\left(\dfrac12\right)^2$. Using $\int dx/\sqrt{u^2-a^2}=\log|u+\sqrt{u^2-a^2}|+C$ gives the stated equivalent form.

$8+3x-x^2=\dfrac{41}{4}-\left(x-\dfrac32\right)^2$. Therefore the integral is $\sin^{-1}\left(\dfrac{x-3/2}{\sqrt{41}/2}\right)+C=\sin^{-1}\left(\dfrac{2x-3}{\sqrt{41}}\right)+C$.

$(x-a)(x-b)=\left(x-\dfrac{a+b}{2}\right)^2-\left(\dfrac{a-b}{2}\right)^2$. Applying the standard form $\int dx/\sqrt{u^2-c^2}=\log|u+\sqrt{u^2-c^2}|+C$ gives the stated equivalent result.

The derivative of $2x^2+x-3$ is $4x+1$. Therefore $\int \dfrac{4x+1}{\sqrt{2x^2+x-3}}dx=2\sqrt{2x^2+x-3}+C$.

Split $x+2$ as $x+2$. Then $\int x/\sqrt{x^2-1}\,dx=\sqrt{x^2-1}$ and $\int dx/\sqrt{x^2-1}=\log|x+\sqrt{x^2-1}|$. Combine the terms.

Write $5x-2=\dfrac56(6x+2)-\dfrac{11}{3}$. Thus the integral is $\dfrac56\log(3x^2+2x+1)-\dfrac{11}{3}\int \dfrac{dx}{3x^2+2x+1}$. Since $3x^2+2x+1=((3x+1)^2+2)/3$, the remaining integral is $\dfrac1{\sqrt2}\tan^{-1}\left(\dfrac{3x+1}{\sqrt2}\right)$.

Let $R=(x-5)(x-4)=x^2-9x+20$. Since $R'=2x-9$ and $6x+7=3(2x-9)+34$, the integral is $6\sqrt{R}+34\int dx/\sqrt{R}$. Using the standard logarithmic form for $\int dx/\sqrt{(x-5)(x-4)}$ gives the answer.

Since $4x-x^2=4-(x-2)^2$ and $x+2=(x-2)+4$, the integral is $\int \dfrac{x-2}{\sqrt{4-(x-2)^2}}dx+4\int \dfrac{dx}{\sqrt{4-(x-2)^2}}$. These give $-\sqrt{4x-x^2}$ and $4\sin^{-1}\left(\dfrac{x-2}{2}\right)$.

Write $x^2+2x+3=(x+1)^2+2$ and $x+2=(x+1)+1$. The first part integrates to $\sqrt{x^2+2x+3}$ and the second to $\log|x+1+\sqrt{x^2+2x+3}|$.

Write $x+3=\dfrac12(2x-2)+4$. Hence the integral is $\dfrac12\log|x^2-2x-5|+4\int \dfrac{dx}{(x-1)^2-6}$. Using $\int dx/(u^2-a^2)=\dfrac1{2a}\log\left|\dfrac{u-a}{u+a}\right|$ gives the result.

Write $x^2+4x+10=(x+2)^2+6$ and $5x+3=\dfrac52(2x+4)-7$. The derivative part gives $5\sqrt{x^2+4x+10}$, and the remaining standard integral gives $-7\log|x+2+\sqrt{x^2+4x+10}|$.

1. i. $x\tan^{-1}(x+1)+C$
2. ii. $\tan^{-1}(x+1)+C$
3. iii. $(x+1)\tan^{-1}x+C$
4. iv. $\tan^{-1}x+C$

$x^2+2x+2=(x+1)^2+1$. Hence the integral is $\tan^{-1}(x+1)+C$, option (ii).

1. i. $\dfrac19\sin^{-1}\left(\dfrac{9x-8}{8}\right)+C$
2. ii. $\dfrac12\sin^{-1}\left(\dfrac{8x-9}{9}\right)+C$
3. iii. $\dfrac13\sin^{-1}\left(\dfrac{9x-8}{8}\right)+C$
4. iv. $\dfrac12\sin^{-1}\left(\dfrac{9x-8}{9}\right)+C$

Complete the square: $9x-4x^2=\dfrac{81}{16}\left[1-\left(\dfrac{8x-9}{9}\right)^2\right]$. Put $u=(8x-9)/9$, so $dx=9du/8$. The integral becomes $\dfrac12\sin^{-1}u+C$, giving option (ii).

Write $\dfrac{x}{(x+1)(x+2)}=\dfrac{-1}{x+1}+\dfrac{2}{x+2}$. Integrating gives $-\log|x+1|+2\log|x+2|+C$.

Since $x^2-9=(x-3)(x+3)$, $\dfrac1{x^2-9}=\dfrac16\left(\dfrac1{x-3}-\dfrac1{x+3}\right)$. Integrating gives the answer.

Decompose $\dfrac{3x-1}{(x-1)(x-2)(x-3)}=\dfrac1{x-1}-\dfrac5{x-2}+\dfrac4{x-3}$. Termwise integration gives the result.

Partial fractions give $\dfrac{x}{(x-1)(x-2)(x-3)}=\dfrac{1}{2(x-1)}-\dfrac2{x-2}+\dfrac{3}{2(x-3)}$. Integrate each term.

Since $x^2+3x+2=(x+1)(x+2)$, $\dfrac{2x}{(x+1)(x+2)}=\dfrac{-2}{x+1}+\dfrac4{x+2}$. Integrating gives the answer.

Divide and decompose: $\dfrac{1-x^2}{x(1-2x)}=\dfrac12+\dfrac1x+\dfrac{3/2}{1-2x}$. Integrating gives $x/2+\og|x|-3\log|1-2x|/4+C$.

Use $\dfrac{x}{(x^2+1)(x-1)}=\dfrac{1}{2(x-1)}+\dfrac{-x+1}{2(x^2+1)}$. Integrating the three pieces gives the stated expression.

Decompose $\dfrac{x}{(x-1)^2(x+2)}=\dfrac{2}{9(x-1)}+\dfrac{1}{3(x-1)^2}-\dfrac{2}{9(x+2)}$. Integrating yields the result.

Since $x^3-x^2-x+1=(x-1)^2(x+1)$, partial fractions give $\dfrac{3x+5}{(x-1)^2(x+1)}=\dfrac{1}{2(x+1)}-\dfrac{1}{2(x-1)}+\dfrac4{(x-1)^2}$. Integrating gives the answer.

Partial fractions give $\dfrac{2x-3}{(x-1)(x+1)(2x+3)}=-\dfrac{1}{10(x-1)}+\dfrac{5}{2(x+1)}-\dfrac{24}{5(2x+3)}$. Integrating gives the stated result.

Since $x^2-4=(x-2)(x+2)$, decompose as $\dfrac{5x}{(x+1)(x-2)(x+2)}=\dfrac{5}{3(x+1)}+\dfrac{5}{6(x-2)}-\dfrac{5}{2(x+2)}$. Integrate termwise.

Divide first: $\dfrac{x^3+x+1}{x^2-1}=x+\dfrac{2x+1}{x^2-1}$. Also $\dfrac{2x+1}{(x-1)(x+1)}=\dfrac{3}{2(x-1)}+\dfrac{1}{2(x+1)}$. Integrating gives the answer.

Write $\dfrac{2}{(1-x)(1+x^2)}=\dfrac1{1-x}+\dfrac{x+1}{1+x^2}$. Integrating gives $-\log|1-x|+\dfrac12\log(1+x^2)+\tan^{-1}x+C$.

Put $u=x+2$. Then $3x-1=3u-7$, so the integrand is $3/u-7/u^2$. Its integral is $3\log|u|+7/u+C$.

$\dfrac1{x^4-1}=\dfrac12\left(\dfrac1{x^2-1}-\dfrac1{x^2+1}\right)$. Integrating gives $\dfrac14\log\left|\dfrac{x-1}{x+1}\right|-\dfrac12\tan^{-1}x+C$.

Multiplying numerator and denominator by $x^{n-1}$ gives $\int \dfrac{x^{n-1}}{x^n(x^n+1)}dx$. Put $t=x^n$, so $dt=nx^{n-1}dx$. The integral is $\dfrac1n\int \dfrac{dt}{t(t+1)}=\dfrac1n\log\left|\dfrac{t}{t+1}\right|+C$.

Put $t=\sin x$, so $dt=\cos x\,dx$. Now $\dfrac1{(1-t)(2-t)}=\dfrac1{1-t}-\dfrac1{2-t}$. Integrating gives $-\log|1-t|+\log|2-t|+C$.

Put $y=x^2$. Then $\dfrac{(y+1)(y+2)}{(y+3)(y+4)}=1+\dfrac{2}{y+3}-\dfrac6{y+4}$. Hence the integrand is $1+\dfrac2{x^2+3}-\dfrac6{x^2+4}$, which integrates to the stated expression.

Put $u=x^2$, so $du=2x\,dx$. Then $\int \dfrac{du}{(u+1)(u+3)}=\dfrac12\log\left|\dfrac{u+1}{u+3}\right|+C$.

Decompose $\dfrac1{x(x^4-1)}=-\dfrac1x+\dfrac{x}{2(x^2-1)}+\dfrac{x}{2(x^2+1)}$. Integrating gives $-\log|x|+\dfrac14\log|x^2-1|+\dfrac14\log(x^2+1)+C=\dfrac14\log|x^4-1|-\log|x|+C$.

Put $t=e^x$, so $dx=dt/t$. Then the integral becomes $\int \dfrac{dt}{t(t-1)}=\log|t-1|-\log|t|+C=\log|e^x-1|-x+C$.

1. i. $\log\left|\dfrac{(x-1)^2}{x-2}\right|+C$
2. ii. $\log\left|\dfrac{(x-2)^2}{x-1}\right|+C$
3. iii. $\log\left|\dfrac{x-1}{x-2}\right|^2+C$
4. iv. $\log|(x-1)(x-2)|+C$

$\dfrac{x}{(x-1)(x-2)}=-\dfrac1{x-1}+\dfrac2{x-2}$. Hence the integral is $-\log|x-1|+2\log|x-2|=\log\left|\dfrac{(x-2)^2}{x-1}\right|+C$.

1. i. $\log|x|-\dfrac12\log(x^2+1)+C$
2. ii. $\log|x|+\dfrac12\log(x^2+1)+C$
3. iii. $-\log|x|+\dfrac12\log(x^2+1)+C$
4. iv. $-\dfrac12\log|x|+\log(x^2+1)+C$

$\dfrac1{x(x^2+1)}=\dfrac1x-\dfrac{x}{x^2+1}$. Integrating gives $\log|x|-\dfrac12\log(x^2+1)+C$, option (i).

Using integration by parts with first function $x$ and second function $\sin x$, $\int x\sin x\,dx=-x\cos x+\int \cos x\,dx=-x\cos x+\sin x+C$.

Take $x$ first. Then $\int x\sin3x\,dx=-\dfrac{x\cos3x}{3}+\dfrac13\int \cos3x\,dx=-\dfrac{x\cos3x}{3}+\dfrac{\sin3x}{9}+C$.

Integrating by parts twice, $\int x^2e^x dx=x^2e^x-2\int xe^x dx=x^2e^x-2e^x(x-1)+C=e^x(x^2-2x+2)+C$.

Take $\log x$ first and $x$ second. Then $\int x\log x\,dx=\dfrac{x^2}{2}\log x-\int \dfrac{x^2}{2}\cdot\dfrac1x dx=\dfrac{x^2}{2}\log x-\dfrac{x^2}{4}+C$.

As in Q4, take $\log2x$ first. Since $d(\log2x)/dx=1/x$, $\int x\log2x\,dx=\dfrac{x^2}{2}\log2x-\dfrac{x^2}{4}+C$.

Take $\log x$ first. Then $\int x^2\log x\,dx=\dfrac{x^3}{3}\log x-\int \dfrac{x^3}{3}\cdot\dfrac1x dx=\dfrac{x^3}{3}\log x-\dfrac{x^3}{9}+C$.

By parts, $I=\dfrac{x^2}{2}\sin^{-1}x-\dfrac12\int \dfrac{x^2}{\sqrt{1-x^2}}dx$. Since $\int \dfrac{x^2}{\sqrt{1-x^2}}dx=\dfrac12\sin^{-1}x-\dfrac{x\sqrt{1-x^2}}{2}$, substitution gives the answer.

By parts, $I=\dfrac{x^2}{2}\tan^{-1}x-\dfrac12\int \dfrac{x^2}{1+x^2}dx$. Since $x^2/(1+x^2)=1-1/(1+x^2)$, $I=\dfrac{x^2+1}{2}\tan^{-1}x-\dfrac{x}{2}+C$.

By parts, $I=\dfrac{x^2}{2}\cos^{-1}x+\dfrac12\int \dfrac{x^2}{\sqrt{1-x^2}}dx$. Use $\int \dfrac{x^2}{\sqrt{1-x^2}}dx=\dfrac12\sin^{-1}x-\dfrac{x\sqrt{1-x^2}}{2}$.

Take $(\sin^{-1}x)^2$ first and $1$ second. Then $I=x(\sin^{-1}x)^2-2\int \dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}dx$. Integrating the remaining term by putting $u=\sqrt{1-x^2}$ gives $I=x(\sin^{-1}x)^2+2\sqrt{1-x^2}\sin^{-1}x-2x+C$.

Put $x=\cos t$. Then $t=\cos^{-1}x$, $dx=-\sin t\,dt$, and $\sqrt{1-x^2}=\sin t$. The integral becomes $-\int t\cos t\,dt=-t\sin t-\os t+C$, i.e. $-\sqrt{1-x^2}\cos^{-1}x-x+C$.

By parts, $\int x\sec^2x\,dx=x\tan x-\int \tan x\,dx=x\tan x+\log|\cos x|+C$.

Take $\tan^{-1}x$ first and $1$ second. Then $\int \tan^{-1}x\,dx=x\tan^{-1}x-\int \dfrac{x}{1+x^2}dx=x\tan^{-1}x-\dfrac12\log(1+x^2)+C$.

By parts with $(\log x)^2$ first, $I=\dfrac{x^2}{2}(\log x)^2-\int x\log x\,dx$. Using Q4 for $\int x\log x\,dx$ gives the result.

Take $\log x$ first and $x^2+1$ second. Then $I=\left(\dfrac{x^3}{3}+x\right)\log x-\int\left(\dfrac{x^3}{3}+x\right)\dfrac{dx}{x}$, giving the answer.

This is of the form $e^x[f(x)+f'(x)]$ with $f(x)=\sin x$. Therefore the integral is $e^x\sin x+C$.

Let $f(x)=1/(1+x)$. Then $f'(x)=-1/(1+x)^2$ and $f+f'=x/(1+x)^2$. Hence the integrand is $e^x[f+f']$, so the integral is $e^xf(x)+C=\dfrac{e^x}{1+x}+C$.

Let $f(x)=\dfrac{\sin x}{1+\cos x}$. Then $f'(x)=\dfrac1{1+\cos x}$ and $f+f'=\dfrac{1+\sin x}{1+\cos x}$. Thus the integral is $e^xf(x)+C$.

Let $f(x)=1/x$. Then $f'(x)=-1/x^2$, so the integrand is $e^x[f(x)+f'(x)]$. Therefore the integral is $e^x/x+C$.

Let $f(x)=1/(x-1)^2$. Then $f'(x)=-2/(x-1)^3$ and $f+f'=(x-3)/(x-1)^3$. Hence the integral is $e^x/(x-1)^2+C$.

Using the standard result obtained by integrating by parts twice, $\int e^{ax}\sin bx\,dx=\dfrac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}+C$. With $a=2,b=1$, the answer follows.

Using $\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)=2\tan^{-1}x$ in the principal interval, the integral is $2\int \tan^{-1}x\,dx=2x\tan^{-1}x-\og(1+x^2)+C$.

1. i. $\dfrac13e^{x^3}+C$
2. ii. $\dfrac13e^{x^2}+C$
3. iii. $\dfrac12e^{x^3}+C$
4. iv. $\dfrac12e^{x^2}+C$

Put $u=x^3$. Then $du=3x^2dx$, so the integral is $\dfrac13e^{x^3}+C$. Hence option (i) is correct.

1. i. $e^x\cos x+C$
2. ii. $e^x\sec x+C$
3. iii. $e^x\sin x+C$
4. iv. $e^x\tan x+C$

Since $\dfrac{d}{dx}(e^x\sec x)=e^x\sec x+e^x\sec x\tan x=e^x\sec x(1+\an x)$, the correct option is (ii).

Use $\int\sqrt{a^2-x^2}\,dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}(x/a)+C$ with $a=2$.

Put $u=2x$, so $dx=du/2$. Then apply the formula for $\int\sqrt{1-u^2}\,du$ and substitute $u=2x$.

Complete the square: $x^2+4x+6=(x+2)^2+2$. Use $\int\sqrt{u^2+a^2}\,du=\dfrac{u}{2}\sqrt{u^2+a^2}+\dfrac{a^2}{2}\log|u+\sqrt{u^2+a^2}|+C$.

$x^2+4x+1=(x+2)^2-3$. Apply $\int\sqrt{u^2-a^2}\,du=\dfrac{u}{2}\sqrt{u^2-a^2}-\dfrac{a^2}{2}\log|u+\sqrt{u^2-a^2}|+C$.

Write $1-x-4x^2=\dfrac{17-(8x+1)^2}{16}$. Put $u=8x+1$, so $dx=du/8$. The integral becomes $\dfrac1{32}\int\sqrt{17-u^2}\,du$, which gives the stated result.

$x^2+4x-5=(x+2)^2-9$. Apply the standard integral of $\sqrt{u^2-a^2}$ with $u=x+2$ and $a=3$.

$x^2+x-3=\left(x+\dfrac12\right)^2-\dfrac{13}{4}$. Applying the $\sqrt{u^2-a^2}$ formula and absorbing constant logarithms gives the answer.

$x^2+3x=\left(x+\dfrac32\right)^2-\dfrac94$. Use the standard $\sqrt{u^2-a^2}$ integral and simplify.

Use the standard formula for $\int\sqrt{x^2+a^2}\,dx$ with $a=1/3$.

1. i. $\dfrac{x}{2}\sqrt{x^2+1}+\dfrac12\log|x+\sqrt{x^2+1}|+C$
2. ii. $\dfrac23(1+x^2)^{3/2}+C$
3. iii. $\dfrac{2}{3x}(1+x^2)^{3/2}+C$
4. iv. $\dfrac{x}{2}\sqrt{x^2+1}+\dfrac{x^2}{2}\log|x+\sqrt{x^2+1}|+C$

This is the standard formula for $\int\sqrt{x^2+a^2}\,dx$ with $a=1$, so option (i) is correct.

1. i. $\dfrac12(x-4)\sqrt{x^2-8x+7}+9\log|x-4+\sqrt{x^2-8x+7}|+C$
2. ii. $\dfrac12(x+4)\sqrt{x^2-8x+7}+9\log|x+4+\sqrt{x^2-8x+7}|+C$
3. iii. $\dfrac12(x-4)\sqrt{x^2-8x+7}-3\log|x-4+\sqrt{x^2-8x+7}|+C$
4. iv. $\dfrac12(x-4)\sqrt{x^2-8x+7}-\dfrac92\log|x-4+\sqrt{x^2-8x+7}|+C$

Complete the square: $x^2-8x+7=(x-4)^2-9$. Applying the standard $\sqrt{u^2-a^2}$ formula with $u=x-4$ and $a=3$ gives option (iv).

An antiderivative is $x^2/2+x$. Thus $\int_{-1}^{1}(x+1)dx=\left[\dfrac{x^2}{2}+x\right]_{-1}^{1}=\dfrac32-\left(-\dfrac12\right)=2$.

$\int_2^3\dfrac{dx}{x}=[\log x]_2^3=\log3-\og2=\log\dfrac32$.

An antiderivative is $x^4-\dfrac53x^3+3x^2+9x$. Evaluating from $1$ to $2$ gives $\dfrac{98}{3}-\dfrac{34}{3}=\dfrac{64}{3}$.

$\int\sin2x\,dx=-\dfrac12\cos2x$. Hence the value is $\left[-\dfrac12\cos2x\right]_{0}^{\pi/4}=0-(-1/2)=1/2$.

$\int\cos2x\,dx=\dfrac12\sin2x$, so $\left[\dfrac12\sin2x\right]_0^{\pi/2}=0$.

$\int_4^5e^x dx=[e^x]_4^5=e^5-e^4$.

$\int\tan x\,dx=\log|\sec x|$. Thus the value is $\log(\sec\pi/4)-\log1=\log\sqrt2=\dfrac12\log2$.

$\int\cosec x\,dx=\log|\cosec x-\cot x|$. Substitution of the limits gives $\log|\sqrt2-1|-\log|2-\sqrt3|=\log\left|\dfrac{\sqrt2-1}{2-\sqrt3}\right|$.

$\int dx/\sqrt{1-x^2}=\sin^{-1}x$. Hence the value is $\sin^{-1}1-\in^{-1}0=\pi/2$.

$\int_0^1\dfrac{dx}{1+x^2}=[\tan^{-1}x]_0^1=\pi/4$.

$\int\dfrac{dx}{x^2-1}=\dfrac12\log\left|\dfrac{x-1}{x+1}\right|$. From $2$ to $3$, this gives $\dfrac12\log\dfrac12-\dfrac12\log\dfrac13=\dfrac12\log\dfrac32$.

Use $\cos^2x=(1+\os2x)/2$. Then the value is $\left[\dfrac{x}{2}+\dfrac{\sin2x}{4}\right]_0^{\pi/2}=\pi/4$.

$\int x/(x^2+1)dx=\dfrac12\log(x^2+1)$. Hence the value is $\dfrac12[\log10-\og5]=\dfrac12\log2$.

Split the integral as $\int_0^1\dfrac{2x}{5x^2+1}dx+3\int_0^1\dfrac{dx}{5x^2+1}$. These are $\dfrac15\log6$ and $\dfrac3{\sqrt5}\tan^{-1}\sqrt5$, respectively.

Put $u=x^2$, so $du=2x\,dx$. Then the value is $\dfrac12[e^u]_0^1=\dfrac{e-1}{2}$.

Divide and decompose: $\dfrac{5x^2}{x^2+4x+3}=5+\dfrac{5}{2(x+1)}-\dfrac{45}{2(x+3)}$. Evaluating from $1$ to $2$ gives $5+\dfrac52\log(3/2)-\dfrac{45}{2}\log(5/4)$.

An antiderivative is $2\tan x+x^4/4+2x$. At $x=\pi/4$ it gives $2+\pi^4/1024+\pi/2$, and at $0$ it gives $0$.

$\sin^2(x/2)-\cos^2(x/2)=-\cos x$. Hence the integral is $[-\sin x]_0^{\pi}=0$.

Split as $3\int_0^2\dfrac{2x}{x^2+4}dx+3\int_0^2\dfrac{dx}{x^2+4}$. These give $3\log2$ and $3\pi/8$, respectively.

$\int_0^1xe^x dx=[(x-1)e^x]_0^1=1$. Also $\int_0^1\sin(\pi x/4)dx=\dfrac4\pi(1-\cos\pi/4)=\dfrac4\pi\left(1-\dfrac1{\sqrt2}\right)$.

1. i. $\dfrac{\pi}{3}$
2. ii. $\dfrac{2\pi}{3}$
3. iii. $\dfrac{\pi}{6}$
4. iv. $\dfrac{\pi}{12}$

The value is $[\tan^{-1}x]_{1}^{\sqrt3}=\pi/3-\pi/4=\pi/12$, so option (iv) is correct.

1. i. $\dfrac{\pi}{6}$
2. ii. $\dfrac{\pi}{12}$
3. iii. $\dfrac{\pi}{24}$
4. iv. $\dfrac{\pi}{4}$

$\int \dfrac{dx}{4+9x^2}=\dfrac16\tan^{-1}\left(\dfrac{3x}{2}\right)$. From $0$ to $2/3$ this is $\dfrac16\tan^{-1}1=\pi/24$, option (iii).

Put $u=x^2+1$, so $du=2x\,dx$. The limits change from $u=1$ to $u=2$. Hence the integral is $\dfrac12\int_1^2du/u=\dfrac12\log2$.

Put $u=\in\phi$, so $du=\os\phi\,d\phi$ and $\cos^4\phi=(1-u^2)^2$. The integral becomes $\int_0^1u^{1/2}(1-u^2)^2du=\left[\dfrac23u^{3/2}-\dfrac47u^{7/2}+\dfrac2{11}u^{11/2}\right]_0^1=\dfrac{64}{231}$.

On $0\le x\le1$, $\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)=2\tan^{-1}x$. Therefore the integral is $2\int_0^1\tan^{-1}x\,dx=[2x\tan^{-1}x-\og(1+x^2)]_0^1=\dfrac{\pi}{2}-\log2$.

Put $u=x+2$, so $x=u-2$ and limits are $2$ to $4$. Then $\int_0^2x\sqrt{x+2}dx=\int_2^4(u^{3/2}-2u^{1/2})du=\left[\dfrac25u^{5/2}-\dfrac43u^{3/2}\right]_2^4=\dfrac{32+16\sqrt2}{15}$.

Put $u=\os x$, so $du=-\sin x\,dx$. The limits change from $1$ to $0$, hence the integral is $\int_0^1\dfrac{du}{1+u^2}=\tan^{-1}1=\pi/4$.

Complete the square: $x+4-x^2=\dfrac{17}{4}-\left(x-\dfrac12\right)^2$. Using $u=x-1/2$ and $a=\sqrt{17}/2$, the antiderivative is $\dfrac1{\sqrt{17}}\log\left|\dfrac{\sqrt{17}+2x-1}{\sqrt{17}-2x+1}\right|$. Evaluating from $0$ to $2$ gives the answer.

$x^2+2x+5=(x+1)^2+4$. Put $u=(x+1)/2$, so $dx=2du$ and the limits are $0$ to $1$. The integral is $\dfrac12\int_0^1\dfrac{du}{1+u^2}=\pi/8$.

Observe that $\dfrac{d}{dx}\left(\dfrac{e^{2x}}{2x}\right)=e^{2x}\left(\dfrac1x-\dfrac{1}{2x^2}\right)$. Therefore the definite integral is $\left[\dfrac{e^{2x}}{2x}\right]_1^2=\dfrac{e^4}{4}-\dfrac{e^2}{2}$.

1. i. $6$
2. ii. $0$
3. iii. $3$
4. iv. $4$

Since $x-x^3=x^3(x^{-2}-1)$, the integrand is $x^{-3}(x^{-2}-1)^{1/3}$. Put $t=x^{-2}-1$, so $dt=-2x^{-3}dx$. The limits change from $8$ to $0$, giving $\dfrac12\int_0^8t^{1/3}dt=6$.

1. i. $\cos x+x\sin x$
2. ii. $x\sin x$
3. iii. $x\cos x$
4. iv. $\sin x+x\cos x$

By the first fundamental theorem of calculus, $\dfrac{d}{dx}\int_0^x t\sin t\,dt=x\sin x$. Hence option (ii) is correct.

Using $\cos^2x=(1+\cos2x)/2$, the integral is $\left[\dfrac{x}{2}+\dfrac{\sin2x}{4}\right]_0^{\pi/2}=\pi/4$.

Let the integral be $I$. Replacing $x$ by $\pi/2-x$ gives $I=\int_0^{\pi/2}\dfrac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$. Adding, $2I=\int_0^{\pi/2}1\,dx=\pi/2$, so $I=\pi/4$.

Replacing $x$ by $\pi/2-x$ changes the numerator to $\cos^{3/2}x$. Adding the two equal transformed integrals gives $2I=\int_0^{\pi/2}1\,dx=\pi/2$. Hence $I=\pi/4$.

Under $x\mapsto\pi/2-x$, the numerator becomes $\sin^5x$. Adding the original and transformed integrals gives $2I=\pi/2$, hence $I=\pi/4$.

Break at $x=-2$: $\int_{-5}^{-2}-(x+2)dx+\int_{-2}^{5}(x+2)dx=\dfrac92+\dfrac{49}{2}=29$.

The graph is symmetric about $x=5$. Thus the value is twice the area of a triangle of base $3$ and height $3$: $2\cdot\dfrac12\cdot3\cdot3=9$.

Put $u=1-x$. Then the integral is $\int_0^1(1-u)u^n du=\dfrac1{n+1}-\dfrac1{n+2}=\dfrac{1}{(n+1)(n+2)}$.

Let $I$ be the integral. Replacing $x$ by $\pi/4-x$ gives $I=\int_0^{\pi/4}\log\dfrac{2}{1+\tan x}dx$. Adding, $2I=\int_0^{\pi/4}\log2\,dx=\dfrac{\pi}{4}\log2$.

Put $u=2-x$. Then $x=2-u$ and the integral becomes $\int_0^2(2-u)\sqrt{u}\,du=2\cdot\dfrac23(2)^{3/2}-\dfrac25(2)^{5/2}=\dfrac{16\sqrt2}{15}$.

Since $\log\sin2x=\log2+\log\sin x+\log\cos x$, the integrand is $\log\sin x-\log\cos x-\og2$. The first two integrals over $[0,\pi/2]$ are equal, so the value is $-(\pi/2)\og2$.

$\sin^2x$ is even, so the integral is $2\int_0^{\pi/2}\sin^2x\,dx=2\cdot\pi/4=\pi/2$.

Let $I$ be the integral. By replacing $x$ with $\pi-x$, $I=\int_0^\pi\dfrac{\pi-x}{1+\sin x}dx$. Adding, $2I=\pi\int_0^\pi\dfrac{dx}{1+\in x}$. With $t=\tan(x/2)$, this last integral equals $2$, hence $I=\pi$.

$\sin^7x$ is an odd function and the limits are symmetric about $0$. Therefore the integral is $0$.

Over a full period, positive and negative contributions of the odd power $\cos^5x$ cancel. Equivalently, split into two intervals differing by $\pi$, where $\cos(x+\pi)=-\cos x$.

Under $x\mapsto\pi/2-x$, the numerator changes sign while the denominator remains the same. Hence the transformed integral is $-I$, but it is also equal to $I$, so $I=0$.

Use $1+\os x=2\cos^2(x/2)$. Then the integral is $\pi\log2+2\int_0^\pi\log|\cos(x/2)|dx=\pi\log2+4\int_0^{\pi/2}\log\cos u\,du=-\pi\log2$.

Let the integral be $I$. Replacing $x$ by $a-x$ gives $I=\int_0^a\dfrac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx$. Adding, $2I=\int_0^a1\,dx=a$, so $I=a/2$.

Break at $x=1$: $\int_0^1(1-x)dx+\int_1^4(x-1)dx=\dfrac12+\dfrac92=5$.

Let $I=\int_0^a f(x)g(x)dx$. Using $x\mapsto a-x$, $I=\int_0^a f(a-x)g(a-x)dx=\int_0^a f(x)g(a-x)dx$. Adding the two forms, $2I=\int_0^a f(x)[g(x)+g(a-x)]dx=4\int_0^a f(x)dx$. Hence $I=2\int_0^a f(x)dx$.

1. i. $0$
2. ii. $2$
3. iii. $\pi$
4. iv. $1$

The functions $x^3$, $x\cos x$, and $\tan^5x$ are odd, so their integrals over symmetric limits are zero. Only the constant $1$ remains, giving length $\pi$. Hence option (iii).

1. i. $2$
2. ii. $\dfrac34$
3. iii. $0$
4. iv. $-2$

Under $x\mapsto\pi/2-x$, the logarithm changes sign. Hence the integral equals its negative, so it is $0$. Therefore option (iii) is correct.

Now $\dfrac{1}{x-x^3}=\dfrac{1}{x(1-x)(1+x)}=\dfrac{1}{x}+\dfrac{1}{2(1-x)}-\dfrac{1}{2(1+x)}$. Hence the integral is $\log|x|-\dfrac12\log|1-x|-\dfrac12\log|1+x|+C=\log|x|-\dfrac12\log|1-x^2|+C$.

Rationalising, $\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}=\dfrac{\sqrt{x+b}-\sqrt{x+a}}{b-a}$. Therefore the integral is $\dfrac1{b-a}\int\left((x+b)^{1/2}-(x+a)^{1/2}\right)dx=\dfrac{2}{3(b-a)}\left((x+b)^{3/2}-(x+a)^{3/2}\right)+C$.

Put $t=\dfrac{a}{x}$, so $x=\dfrac{a}{t}$ and $dx=-\dfrac{a}{t^2}\,dt$. Also $x\sqrt{ax-x^2}=\dfrac{a^2\sqrt{t-1}}{t^2}$. The integral becomes $-\dfrac1a\int(t-1)^{-1/2}\,dt=-\dfrac{2}{a}\sqrt{t-1}+C=-\dfrac{2}{a}\sqrt{\dfrac{a-x}{x}}+C$.

Let $F=\dfrac{(x^4+1)^{1/4}}{x}$. Then $F'=\dfrac{x^3(x^4+1)^{-3/4}\cdot x-(x^4+1)^{1/4}}{x^2}=\dfrac{x^4-(x^4+1)}{x^2(x^4+1)^{3/4}}=-\dfrac1{x^2(x^4+1)^{3/4}}$. Hence the required integral is $-F+C$.

Put $x=t^6$, so $dx=6t^5\,dt$. Then the integral becomes $\int\dfrac{6t^5}{t^3+t^2}\,dt=6\int\dfrac{t^3}{t+1}\,dt=6\int\left(t^2-t+1-\dfrac1{t+1}\right)dt$. Thus it equals $2t^3-3t^2+6t-6\log|t+1|+C$. Substituting $t=x^{1/6}$ gives the answer.

Write $\dfrac{5x}{(x+1)(x^2+9)}=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+9}$. Comparing coefficients gives $A=-\dfrac12$, $B=\dfrac12$, $C=\dfrac92$. Hence the integral is $-\dfrac12\log|x+1|+\dfrac12\int\dfrac{x}{x^2+9}\,dx+\dfrac92\int\dfrac{dx}{x^2+9}$, which simplifies to the answer.

Use $\sin x=\sin(x-a)\cos a+\cos(x-a)\sin a$. Then $\dfrac{\sin x}{\sin(x-a)}=\cos a+\sin a\cot(x-a)$. Therefore the integral is $x\cos a+\sin a\log|\sin(x-a)|+C$.

Since $e^{k\log x}=x^k$, the integrand is $\dfrac{x^5-x^4}{x^3-x^2}=\dfrac{x^4(x-1)}{x^2(x-1)}=x^2$. Hence the integral is $\int x^2\,dx=\dfrac{x^3}{3}+C$.

Put $u=\sin x$, so $du=\cos x\,dx$. The integral becomes $\int\dfrac{du}{\sqrt{4-u^2}}=\sin^{-1}\left(\dfrac{u}{2}\right)+C=\sin^{-1}\left(\dfrac{\sin x}{2}\right)+C$.

Now $\sin^8x-\cos^8x=(\sin^4x-\cos^4x)(\sin^4x+\cos^4x)=-(\cos2x)(1-2\sin^2x\cos^2x)$. Hence the integrand is $-\cos2x$, and the integral is $-\dfrac12\sin2x+C$.

Since $\tan(x+b)-\tan(x+a)=\dfrac{\sin(b-a)}{\cos(x+a)\cos(x+b)}$, the integrand is $\dfrac{\tan(x+b)-\tan(x+a)}{\sin(b-a)}$. Integrating gives $\dfrac1{\sin(b-a)}\left(\log|\cos(x+a)|-\log|\cos(x+b)|\right)+C$, which is the answer.

Put $u=x^4$, so $du=4x^3\,dx$. The integral is $\dfrac14\int\dfrac{du}{\sqrt{1-u^2}}=\dfrac14\sin^{-1}u+C=\dfrac14\sin^{-1}(x^4)+C$.

Put $u=e^x$, so $du=e^x\,dx$. Then $\int\dfrac{du}{(u+1)(u+2)}=\int\left(\dfrac1{u+1}-\dfrac1{u+2}\right)du=\log\left|\dfrac{u+1}{u+2}\right|+C$. Substituting $u=e^x$ gives the answer.

Now $\dfrac{1}{(x^2+1)(x^2+4)}=\dfrac13\left(\dfrac1{x^2+1}-\dfrac1{x^2+4}\right)$. Therefore the integral is $\dfrac13\tan^{-1}x-\dfrac13\cdot\dfrac12\tan^{-1}\left(\dfrac{x}{2}\right)+C$, i.e. $\dfrac13\tan^{-1}x-\dfrac16\tan^{-1}\left(\dfrac{x}{2}\right)+C$.

Since $e^{\log\sin x}=\sin x$, the integral is $\int\sin x\cos^3x\,dx$. Put $u=\cos x$, so $du=-\sin x\,dx$. Hence the integral is $-\int u^3\,du=-\dfrac{u^4}{4}+C=-\dfrac14\cos^4x+C$.

Here $e^{3\log x}=x^3$. Thus the integral is $\int\dfrac{x^3}{x^4+1}\,dx$. Put $u=x^4+1$, so $du=4x^3\,dx$. Therefore the integral is $\dfrac14\log u+C=\dfrac14\log(x^4+1)+C$.

Put $u=f(ax+b)$. Then $du=a f'(ax+b)\,dx$. Hence $\int f'(ax+b)[f(ax+b)]^n\,dx=\dfrac1a\int u^n\,du=\dfrac{u^{n+1}}{a(n+1)}+C$, for $n\ne -1$. Substituting back gives the answer.

Since $\sin(x+\alpha)=\sin x\cos\alpha+\cos x\sin\alpha=\sin x(\cos\alpha+\sin\alpha\cot x)$, the integrand becomes $\dfrac{\csc^2x}{\sqrt{\cos\alpha+\sin\alpha\cot x}}$. Put $u=\cos\alpha+\sin\alpha\cot x$; then $du=-\sin\alpha\csc^2x\,dx$. The integral is $-\dfrac1{\sin\alpha}\int u^{-1/2}\,du=-\dfrac{2}{\sin\alpha}\sqrt{u}+C$.

Put $u=\sqrt{x}$, so $dx=2u\,du$. Then put $u=\cos\theta$. The integral becomes $2\int u\sqrt{\dfrac{1-u}{1+u}}\,du=-2\int\cos\theta(1-\os\theta)\,d\theta$. This is $-2\sin\theta+\theta+\sin\theta\cos\theta+C$. Since $\theta=\cos^{-1}\sqrt{x}$, $\sin\theta=\sqrt{1-x}$ and $\sin\theta\cos\theta=\sqrt{x(1-x)}$, the answer follows.

Since $1+\cos2x=2\cos^2x$ and $\sin2x=2\sin x\cos x$, the trigonometric factor is $\dfrac{2+\sin2x}{1+\cos2x}=\sec^2x+\an x$. Therefore the integrand is $e^x(\tan x+\sec^2x)$, which is the derivative of $e^x\tan x$. Hence the integral is $e^x\tan x+C$.

Write $\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}+\dfrac{C}{x+2}$. Comparing coefficients gives $A=-2$, $B=1$, $C=3$. Therefore the integral is $-2\log|x+1|+\int\dfrac{dx}{(x+1)^2}+3\log|x+2|+C$, i.e. $-2\log|x+1|-\dfrac1{x+1}+3\log|x+2|+C$.

Use $\tan^{-1}\sqrt{\dfrac{1-x}{1+x}}=\dfrac12\cos^{-1}x$. Hence the integral is $\dfrac12\int\cos^{-1}x\,dx$. By parts, $\int\cos^{-1}x\,dx=x\cos^{-1}x-\sqrt{1-x^2}+C$. Thus the required integral is $\dfrac12\left(x\cos^{-1}x-\sqrt{1-x^2}\right)+C$.

Put $t=\dfrac{\sqrt{x^2+1}}{x}$. Then $\log(x^2+1)-2\log x=\log\dfrac{x^2+1}{x^2}=2\log t$ and $dt=-\dfrac{dx}{x^2\sqrt{x^2+1}}$. Also $t^2=\dfrac{x^2+1}{x^2}$, so the integrand becomes $-2t^2\log t\,dt$. Hence the integral is $-2\left(\dfrac{t^3\log t}{3}-\dfrac{t^3}{9}\right)+C=\left(\dfrac29-\dfrac13\log\dfrac{x^2+1}{x^2}\right)t^3+C$.

Since $\dfrac{1-\sin x}{1-\cos x}=-\cot\dfrac{x}{2}+\dfrac12\csc^2\dfrac{x}{2}$, the integrand is the derivative of $-e^x\cot\dfrac{x}{2}$. Therefore the integral is $\left[-e^x\cot\dfrac{x}{2}\right]_{\pi/2}^{\pi}=0-(-e^{\pi/2})=e^{\pi/2}$.

Put $t=\tan x$. Then $\sin x\cos x\,dx=\dfrac{t}{(1+t^2)^2}dt$ and $\cos^4x+\sin^4x=\dfrac{1+t^4}{(1+t^2)^2}$. The limits are $0$ and $1$. Thus the integral is $\int_0^1\dfrac{t}{1+t^4}\,dt=\dfrac12\left[\tan^{-1}(t^2)\right]_0^1=\dfrac{\pi}{8}$.

Put $t=\tan x$. Then the integral becomes $\int_0^\infty\dfrac{dt}{(1+t^2)(1+4t^2)}$. Since $\dfrac{1}{(1+t^2)(1+4t^2)}=-\dfrac1{3(1+t^2)}+\dfrac4{3(1+4t^2)}$, the value is $-\dfrac13\cdot\dfrac{\pi}{2}+\dfrac43\cdot\dfrac{\pi}{4}=\dfrac{\pi}{6}$.

Put $u=\sin x-\cos x$, so $du=(\sin x+\cos x)\,dx$ and $\sin2x=1-u^2$. At $x=\pi/6$, $u=\dfrac{1-\sqrt3}{2}$; at $x=\pi/3$, $u=\dfrac{\sqrt3-1}{2}$. Hence the integral is $\int_{-(\sqrt3-1)/2}^{(\sqrt3-1)/2}\dfrac{du}{\sqrt{1-u^2}}=2\sin^{-1}\left(\dfrac{\sqrt3-1}{2}\right)$.

Rationalising, $\dfrac{1}{\sqrt{1+x}-\sqrt{x}}=\sqrt{1+x}+\sqrt{x}$. Therefore the integral is $\left[\dfrac{2}{3}(1+x)^{3/2}+\dfrac{2}{3}x^{3/2}\right]_0^1=\dfrac23(2\sqrt2+1)-\dfrac23=\dfrac{4\sqrt2}{3}$.

Put $u=\sin x-\cos x$, so $du=(\sin x+\cos x)\,dx$ and $\sin2x=1-u^2$. The limits are $-1$ and $0$. Thus the integral is $\int_{-1}^{0}\dfrac{du}{25-16u^2}=\dfrac1{40}\left[\log\left|\dfrac{5+4u}{5-4u}\right|\right]_{-1}^{0}=\dfrac1{20}\log3$.

Put $t=\sin x$, so $\sin2x\,dx=2t\,dt$ and the limits are $0$ to $1$. The integral is $2\int_0^1t\tan^{-1}t\,dt$. By parts, this equals $\left[t^2\tan^{-1}t\right]_0^1-\int_0^1\dfrac{t^2}{1+t^2}\,dt=\dfrac\pi4-(1-\dfrac\pi4)=\dfrac\pi2-1$.

Split the interval at $2$ and $3$. On $[1,2]$ the integrand is $(x-1)+(2-x)+(3-x)=4-x$; on $[2,3]$ it is $(x-1)+(x-2)+(3-x)=x$; on $[3,4]$ it is $(x-1)+(x-2)+(x-3)=3x-6$. Hence the value is $\int_1^2(4-x)\,dx+\int_2^3x\,dx+\int_3^4(3x-6)\,dx=\dfrac52+\dfrac52+\dfrac92=\dfrac{19}{2}$.

Using partial fractions, $\dfrac1{x^2(x+1)}=-\dfrac1x+\dfrac1{x^2}+\dfrac1{x+1}$. Therefore $\int_1^3\dfrac{dx}{x^2(x+1)}=\left[-\log x-\dfrac1x+\log(x+1)\right]_1^3$. This equals $\left(-\log3-\dfrac13+\log4\right)-(-1+\log2)=\dfrac23+\log\dfrac{4}{3}-\log2=\dfrac23+\log\dfrac23$.

Integrating by parts, $\int xe^x\,dx=xe^x-\int e^x\,dx=(x-1)e^x+C$. Thus $\int_0^1xe^x\,dx=\left[(x-1)e^x\right]_0^1=0-(-1)=1$.

The function $x^{17}\cos^4x$ is odd because $x^{17}$ is odd and $\cos^4x$ is even. The integral of an odd function over a symmetric interval $[-a,a]$ is $0$. Hence $\int_{-1}^{1}x^{17}\cos^4x\,dx=0$.

Write $\sin^3x=\sin x(1-\os^2x)$. Put $u=\cos x$, so $du=-\sin x\,dx$. Then $\int_0^{\pi/2}\sin^3x\,dx=\int_0^{\pi/2}\sin x(1-\os^2x)\,dx=\int_0^1(1-u^2)\,du=\left[u-\dfrac{u^3}{3}\right]_0^1=\dfrac23$.

Since $\tan^3x=\tan x(\sec^2x-1)$, $2\int\tan^3x\,dx=2\int\tan x\sec^2x\,dx-2\int\tan x\,dx=\tan^2x+2\log|\cos x|+C$. Hence $\int_0^{\pi/4}2\tan^3x\,dx=\left[\tan^2x+2\log(\cos x)\right]_0^{\pi/4}=1+2\log\dfrac1{\sqrt2}=1-\log2$.

By parts, $\int\sin^{-1}x\,dx=x\sin^{-1}x-\int\dfrac{x}{\sqrt{1-x^2}}\,dx=x\sin^{-1}x+\sqrt{1-x^2}+C$. Therefore $\int_0^1\sin^{-1}x\,dx=\left[x\sin^{-1}x+\sqrt{1-x^2}\right]_0^1=\dfrac\pi2-1$.

1. i. $\tan^{-1}(e^x)+C$
2. ii. $\tan^{-1}(e^{-x})+C$
3. iii. $\log(e^x-e^{-x})+C$
4. iv. $\log(e^x+e^{-x})+C$

Multiply numerator and denominator by $e^x$ to get $\int\dfrac{e^x\,dx}{e^{2x}+1}$. Put $u=e^x$, so $du=e^x\,dx$. The integral becomes $\int\dfrac{du}{1+u^2}=\tan^{-1}u+C=\tan^{-1}(e^x)+C$. Therefore option (i) is correct.

1. i. $-\dfrac{1}{\sin x+\cos x}+C$
2. ii. $\log|\sin x+\cos x|+C$
3. iii. $\log|\sin x-\cos x|+C$
4. iv. $\dfrac{1}{(\sin x+\cos x)^2}$

Since $\dfrac{d}{dx}(\sin x+\cos x)=\cos x-\sin x$ and $\cos2x=(\cos x-\sin x)(\cos x+\sin x)$, the integrand is $\dfrac{\cos x-\sin x}{\sin x+\cos x}$. Therefore the integral is $\log|\sin x+\cos x|+C$. Option (ii) is correct.

1. i. $\dfrac{a+b}{2}\int_a^b f(b-x)\,dx$
2. ii. $\dfrac{a+b}{2}\int_a^b f(b+x)\,dx$
3. iii. $\dfrac{b-a}{2}\int_a^b f(x)\,dx$
4. iv. $\dfrac{a+b}{2}\int_a^b f(x)\,dx$

Let $I=\int_a^b x f(x)\,dx$. Using the property $\int_a^b g(x)\,dx=\int_a^b g(a+b-x)\,dx$, $I=\int_a^b(a+b-x)f(a+b-x)\,dx=\int_a^b(a+b-x)f(x)\,dx$. Adding this to the original integral gives $2I=(a+b)\int_a^b f(x)\,dx$. Hence $I=\dfrac{a+b}{2}\int_a^b f(x)\,dx$, so option (iv) is correct.