If $$\displaystyle \int x^{2}e^{-2x}dx=e^{-2x}(ax^{2}+bx+c)+d$$ then

$$a=-\displaystyle \frac {1}{2}$$

$$c=-\displaystyle \frac{1}{4}$$

MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 4

$\displaystyle \int x$ log

$\displaystyle (1+\frac{1}{x})dx =$

$f(x)\log(x+1)+g(x).x^{2}+Lx+c$ , then

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$f(x) =\displaystyle \frac{x^{2}}{2}-\frac{1}{2},\ g(x) =\displaystyle \frac{1}{2}\log x,\ L=1$
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$f(x) =\displaystyle \frac{x^{2}}{2}+\frac{1}{2},g(x)=-\frac{1}{2}\log x,L=\frac{1}{2}$
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$f(x)=\displaystyle \frac{x^{2}}{2}-\frac{1}{2},g(x)=-\frac{1}{2}\log x,L=\frac{1}{2}$
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$f(x) =\displaystyle \frac{x^{2}}{2}-\frac{1}{2},g(x)=\frac{1}{2}\log x,\ L=-\displaystyle \frac{1}{2}$

Explanation

$\displaystyle \int x \log(1+\dfrac {1}{x})dx.$

$\displaystyle =\log (1+\dfrac {1}{x})\cdot \dfrac {x^2}{2}-\int (\dfrac {d }{dx}\log(1+\dfrac {1}{x})\int x dx)\cdot dx$

$\displaystyle =\log (1+\dfrac {1}{x})\cdot \dfrac {x^2}{2}-\int \dfrac {x}{(\dfrac {x+1}{x})}\cdot \dfrac {-1}{x}x dx.$

$\displaystyle =\log (1+\dfrac {1}{x})\cdot \dfrac {x^2}{2}+\int \dfrac {x}{x+1}\cdot dx.$

$\displaystyle =\log (1+\dfrac {1}{x})\cdot \dfrac {x^2}{2}+x-\log (x+1).$

$\displaystyle =[\log (1+x)]\dfrac {x^2}{2}$

$\displaystyle \int x^{2}(\log x)^{3}dx=$

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$\displaystyle x^{3}\left[ \frac { (logx)^{ 3 } }{ 3 } -\frac { (logx)^{ 2 } }{ 3 } +\frac { 2(logx) }{ 9 } -\frac { 2 }{ 27 } \right] +c$
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$\displaystyle x^{3}\left[ \frac { (\log x)^{ 3 } }{ 3 } +\frac { (\log x)^{ 2 } }{ 3 } +\frac { 2(\log x) }{ 9 } +\frac { 2 }{ 27 } \right] +c$
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$\displaystyle x^{3}\left[ \frac { (\log x)^{ 3 } }{ 3 } -\frac { (\log x)^{ 2 } }{ 27 } \right] +c$
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$\displaystyle x^{3}\left[ \frac { (\log x)^{ 3 } }{ 3 } +\frac { (\log x)^{ 2 } }{ 3 } +\frac { 2(\log x) }{ 9 } -\frac { 2 }{ 27 } \right] +c$

Explanation

$\int x^{2}(log\ x)^{3}dx$

$=(log\ x)^{3}\int x^{2}dx$

$= - \int \left ( \dfrac{d}{dx}(log\ x)^{3}.\int x^{2}.dx \right ).dx$

$= (log\ x)^{3}\ \dfrac{x^{3}}{3} \int 3(log\ x)^{2}-\ ^{1}/_{x}.\ ^{x^{2}}/_{3} dx$

$(log\ x)^{3}\ \dfrac{x^{3}}{3}-\int (log\ x)^{2}.dx$

$\int x^{2}(log\ x)^{2}= (log\ x)^{2} \dfrac{x^{3}}{3}- \int \ ^{1}/_{2} \dfrac{log\ x}{x}.\ ^{x^{2}}/_{3}$

$= (log\ x)^{2}-\ ^{x^{3}}/_{3}\ -\ ^{1}/_{6} \int log\ x.x^{2}$

$\int log\ x.x^{2}= log\ x.\ ^{x^{3}}/_{3}- \int \ ^{1}/_{x}\ x^{2}/3\ dx$

$logx-x^{3}/3-x^{3}/q+c$

$=>x^{3}\ [(logx)^{3}-(logx)^{2}+\dfrac{2}{9}logx-\dfrac{2}{27}]+c$

$\displaystyle \int xsin^{2}xdx$

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$\displaystyle \frac{x^{2}}{4}-\frac{\cos 2x}{3}+\frac{1}{8}sin2x +c$
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$\displaystyle \frac{x^{2}}{4}-\frac{xsin2x}{4}-\frac{1}{8}cos2x +c$
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$\displaystyle \frac{x^{2}}{4}+\frac{xsin2x}{4}+\frac{1}{8}cos2x +c$
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$\displaystyle \frac{-x^{2}}{4}-\frac{cos2x}{3}+\frac{1}{8}cos2x +c$

Explanation

$\int x\ sin^{2}x\ dx.$

$= x \int \ sin^{2}x.\ dx$

$-\int \left ( \int sin^{2}x.dx \right )dx$

$cos\ 2x= 1-2\ sin^{2}x$

$2\ sin^{2}x= \frac{1-cos\ 2x}{2}$

$x\ \int \left ( \frac{1- cos\ 2x}{2} \right )dx-\int \left ( \int \left ( \frac{1- cos\ 2x}{2} \right ).dx \right )dx.$

$= \frac{x}{2}[x- \ ^{1}/_{2}sin\ 2x]- \int [\frac{x}{2}- \ ^{1}/_{2}sin\ 2x]dx$

$= \frac{x^{2}}{2}- \frac{x}{4} sin\ 2x- \int [\frac{x^{2}}{4}+ \frac{1}{4} cos\ 2x]$

$= \frac{x^{2}}{4}- \frac{x\ sin\ 2x}{4}- \frac{1}{8} cos\ 2x+c$

$\displaystyle \int\frac{x}{1+\cos x}dx=$

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$x\tan \frac{x}{2}-2\log|\sec x/2|+c$
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$-x\tan x/2-\displaystyle \frac{1}{2}\log|\sec x/2|+c$
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$x\tan x/2+\displaystyle \frac{1}{2}\log|\sec x/2|+c$
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$x\cot x/2-\displaystyle \frac{1}{2}\log|\csc x/2|+c$

Explanation

$\int \dfrac{x}{1+cos\ x}dx.$

$= x \int \dfrac{1}{1+cos\ x}dx- \int 1 \left ( \int \dfrac{1}{1+cos\ x}dx \right )dx$

$\int \dfrac{1}{1+cos\ x}dx= tan \frac{x}{2}$

$x \ tan \dfrac{x}{2}- \int tan \dfrac{x}{2}.dx.$

$x tan \dfrac{x}{2}+ 2\ log \left | cos\ x \right |+c$

$= x\ tan \dfrac{x}{2}- 2\ log \left | sec \dfrac{x}{2} \right |+ c$

Area bounded by

$\mathrm{y}=\{\mathrm{x}\},\{.\}$ is fractional part of function and

$\mathrm{x}=\pm 1$ is in sq. units

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1
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2
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3
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4

Explanation

Area

$\displaystyle =2\int _{ 0 }^{ 1 }{ xdx } =2{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 1 }=2.\frac { 1 }{ 2 } =1$

$\displaystyle \int cos(\log x)dx=$

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$x[cos(\log x)-sin(\log x)]+c$
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$\displaystyle \frac{x}{2}[cos(\log x)-sin(\log x)]+c$
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$\displaystyle \frac{\log x}{2}[cosx+sinx]+c$
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$\dfrac{x}{2}[cos(\log x)+sin(\log x)]+c$

Explanation

$\int cos(log x)^1dx$

$x cos (log x)+\int (\dfrac {sin(log x)}{x}\cdot x)dx$

$=x cos (log x)+\int sin(log x)dx$

$\int sin (log x)dx=x sin (log x)+\int cos (log x)dx$

$\int cos (log x)dx=x cos (log x)+x sin (log x)-\int cos (log x)dx$

$=\int cos (log x)dx=\dfrac {x}{2}[cos (log x)+sin (log x)]+c.$

$\displaystyle \int\frac{\log(\log x)}{x}dx=$

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$\displaystyle \log x[\log(\log x)-1]+c$
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$\displaystyle x[\log(\log x)+1]+c$
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$\displaystyle \log x[\log(\log x)+1]+c$
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$\displaystyle x[\log(\log x)-1]+c$

Explanation

$\int \dfrac{log(logx)}{x}dx$

$= log(log\ x)(log\ x)-\int \left ( \dfrac{1}{log\ x}\dfrac{1}{x}log\ x \right )dx$

$= log(log\ x)[log\ x]-log\ x+c$

$= log\ x [log(log\ x)-1]+c$

$\displaystyle \int 2^{mx}.3^{nx}dx$ when

$m,n\epsilon N$ is equal to:

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$\displaystyle \frac{2^{mx}+3^{nx}}{m \ln 2+n\ln 3}+c$
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$\displaystyle\frac{e^{(mx\ln2+nx\ln3)}}{m\ln2+n\ln3}+c$
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$\displaystyle\frac{2^{mx}.3^{nx}}{\ln(2^{m}\cdot 3^{n})}+c$
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None of these

Explanation

Let,

$\displaystyle I = \int 2^{mx}.3^{nx}dx$
Now put

$2^m=a, 3^n = b$

$\Rightarrow \displaystyle I = \int (a^x.b^x)dx$
Now using by parts, by considering

$a^x$ as 1st function and

$b^x$ as 2nd function

$\Rightarrow \displaystyle I = a^x\frac{b^x}{\ln b}-\int a^x\ln a.\frac{b^x}{\ln b}dx+k$

$\Rightarrow \displaystyle I = \frac{a^xb^x}{\ln b}-\frac{\ln a}{\ln b}\int a^xb^xdx+k$

$\Rightarrow \displaystyle I = \frac{a^xb^x}{\ln b}-\frac{\ln a}{\ln b}I+k$

$\Rightarrow \displaystyle I = \frac{a^xb^x}{\ln a+\ln b}+\frac{k\ln b}{\ln a+\ln b}$

$\Rightarrow \displaystyle I = \frac{a^xb^x}{\ln a+\ln b}+c$ , where

$\displaystyle c=\frac{k\ln b}{\ln a+\ln b}\in$

$R$
Hence option B,C are correct choice

$\underset {x\to \infty}\lim \int_{0}^{x}xe^{t^{2}-x^{2}}dt$ is equal to

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$2$
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$\displaystyle -\frac{1}{2}$
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$\displaystyle \frac{1}{2}$
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$1$

Explanation

$\displaystyle \lim _{ x\to \infty }{ \int _{ 0 }^{ x } xe^{ t^{ 2 }-x^{ 2 } }dt } =\lim _{ x\to \infty }{ \int _{ 0 }^{ x } \dfrac { xe^{ t^{ 2 } } }{ { e }^{ { x }^{ 2 } } } dt } =\lim _{ x\to \infty }{ \int _{ 0 }^{ x } \dfrac { e^{ t^{ 2 } } }{ \dfrac { { e }^{ { x }^{ 2 } } }{ x } } dt }$

$\displaystyle \lim _{ x\to \infty }{ \dfrac { { e }^{ { x }^{ 2 } } }{ \dfrac { x(2x){ e }^{ { x }^{ 2 } }-{ e }^{ { x }^{ 2 } } }{ x^{ 2 } } } } =\lim _{ x\to \infty }{ \dfrac { { e }^{ { x }^{ 2 } } }{ { e }^{ { x }^{ 2 } }\left( \dfrac { 2x^{ 2 }-1 }{ x^{ 2 } } \right) } } =\lim _{ x\to \infty }{ \dfrac { 1 }{ 2-\dfrac { 1 }{ x^{ 2 } } } } =\dfrac { 1 }{ 2 }$

$\displaystyle \int \sin^{-1}\left ( \frac{2x}{1+x^2} \right )dx$ is equal to

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$\displaystyle 2(x \tan^{-1} x +\ln \left | \cos (\tan^{-1}x) \right |)+C$
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$2[(\displaystyle x\tan^{-1}{x})^2+\ln \left | \sec (\tan^{-1}x) \right |]+c$
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$2[(\displaystyle x\tan^{-1}{x})^2-\ln \left | \cos (\tan^{-1}x) \right |]+c$
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None of these

Explanation

$\displaystyle I=\int \sin^{-1}\left ( \frac{2x}{1+x^{2}} \right )dx$
Let

$x=\tan \theta$

$\implies \displaystyle dx=\sec ^{2}\theta d\theta$

$\therefore \displaystyle I=\int \sin^{-1}\left ( \frac{2\tan \theta }{1+\tan^{2}\theta} \right ) \sec^{2}\theta d\theta$

$=\displaystyle \int \sin^{-1}\left(\dfrac{2\frac{\sin \theta}{\cos \theta}}{1+\frac{\sin^{2}\theta}{\cos^{2}\theta}}\right)\sec^{2}\theta d\theta$

$=\displaystyle \int \sin^{-1}\left(\dfrac{2\frac{\sin \theta}{\cos \theta}}{\frac{\sin^{2}\theta+\cos^{2}\theta}{\cos^{2}\theta}}\right)\sec^{2}\theta d\theta$

$=\displaystyle \int \sin^{-1}(2\sin{\theta} \cos{\theta})\sec^{2}\theta d\theta$

$=\displaystyle \int \sin^{-1}(\sin{2\theta})\sec^{2}\theta d\theta$

$=\displaystyle 2\int \theta \sec ^{2}\theta \:d\theta$

$=\displaystyle 2(\theta \:\tan \:\theta +\ln \left | \cos \:\theta \right |)+C$

$=\displaystyle 2(x \tan^{-1} x +\ln \left | \cos (\tan^{-1}x) \right |)+C$

Ans: A

$\displaystyle \int e^{tan^{-1}x}\left[\frac{1+x+x^{2}}{1+x^{2}}\right]dx=$

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$\displaystyle x^{2}e^{\tan^{-1}x}+c$
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$\displaystyle x e^{\tan^{-1}x}+c$
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$\displaystyle e^{\tan^{-1}x}+c$
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$\displaystyle \frac{1}{2}e^{\tan^{-1}x}+c$

Explanation

Let

$I=\displaystyle \int e^{\tan^{-1}x}\left[\frac{1+x+x^{2}}{1+x^{2}}\right]dx$

$=\displaystyle \int e^{\tan^{-1}x}\left[1+\frac {x}{1+x^2}\right]dx$

$=\displaystyle \int \left[e^{\tan^{-1}x}+\frac {xe^{\tan^{-1}x}}{1+x^2}\right]dx$
For

$f(x)=e^{\tan^{-1} x}$ , above integral

$I$ reduces in the form of

$\displaystyle \int [f(x)+x f^{'}(x)]dx$

But we know that,

$\dfrac{d}{dx} (xf(x)+c) = \displaystyle [f(x)+x f^{'}(x)]$

$\therefore \displaystyle \int [f(x)+x f^{'}(x)]dx=xf(x)+c$

$\therefore I = \displaystyle xe^{\tan^{-1}x}+c$

$\therefore I=\displaystyle \int e^{\tan^{-1}x}\left[\frac{1+x+x^{2}}{1+x^{2}}\right]dx=xe^{\tan ^{-1} x}+c$
Hence, option 'B' is correct.

$\displaystyle \int x^{2}e^{-2x}dx=e^{-2x}(ax^{2}+bx+c)+d$ then

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$a=-\displaystyle \frac {1}{2}$
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$b=2$
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$c=-\displaystyle \frac{1}{4}$
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$d\in R$

Explanation

$\displaystyle \int x^{2}e^{-2x}dx=e^{-2x}(ax^{2}+bx+c)+d$
Differentiating both sides, we get

$x^{2}e^{-2x}=e^{-2x}(2ax+b)+(ax^{2}+bx+c)(-2e^{-2x})$

$x^2e^{-2x}=e^{-2x}(-2ax^{2}+2(a-b)x+b-2c)$
On comparing the coefficients, we get

$-2a=1,\:2(a-b)=0,\:b-2c=0$

$\displaystyle a=\frac{-1}{2}, b=-\frac{1}{2},\:c=-\frac{1}{4}$
Also,

$d\in R$

$\displaystyle \int x.\frac{\ln \left ( x+\sqrt{1+x^2} \right )}{\sqrt{1+x^2}} dx$ equals

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$\sqrt{1+x^2}\ln \left ( x+\sqrt{1+x^2} \right )-x+c$
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$\displaystyle \frac{x}{2}.\ln^2\left ( x+\sqrt{1+x^2} \right )-\displaystyle \frac{x}{\sqrt{1+x^2}}+c$
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$\displaystyle \frac{x}{2}.\ln^2\left ( x+\sqrt{1+x^2} \right )+\displaystyle \frac{x}{\sqrt{1+x^2}}+c$
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$\sqrt{1+x^2}\ln \left ( x+\sqrt{1+x^2} \right )+x+c$

Let

$\displaystyle \int e^{x}\left \{ f(x)-f'(x) \right \}dx=\phi (x).$ Then

$\int e^{x} f(x)dx$ is

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$\displaystyle \phi (x)=e^{x}f(x)$
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$\displaystyle \phi (x)-e^{x}f(x)$
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$\displaystyle \frac{1}{2}\left \{ \phi (x)+e^{x}f(x) \right \}$
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$\displaystyle \frac{1}{2}\left \{ \phi (x)+e^{x}f'(x) \right \}$

Explanation

Given

$\displaystyle \int e^{x}\left \{ f(x)-f'(x) \right \}dx=\phi (x).$ ...(1)

Consdier,

$\int e^{ x }\left\{ { f }(x)-{ f }'(x) \right\} dx$

$=\int e^{ x }f(x)dx-\int e^{ x }f'(x)dx$

$=e^{ x }f(x)-2\int e^{ x }f'(x)dx$

So, by (1),

$e^{ x }f(x)-2\int e^{ x }f'(x)dx=\phi (x)$

$\displaystyle \int e^xf(x)dx =\frac{1}{2}\left \{ \phi (x)+e^{x}f(x) \right \}$

$\displaystyle \int e^{\displaystyle \tan x}(\sec x-\sin x)dx$ is equal to

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$\displaystyle e^{\displaystyle \tan x}\cos x+C$
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$\displaystyle e^{\displaystyle \tan x}\sin x+C$
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$\displaystyle -e^{\displaystyle \tan x}\cos x+C$
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$\displaystyle e^{\displaystyle \tan x}\sec x+C$

Explanation

$I=\displaystyle \int e^{\tan x}(\sec x-\sin x)dx$

$=-\displaystyle \int \sin xe^{\tan x}dx+ \int \sec xe^{\tan x}dx$

$=\displaystyle e^{\tan x}\cos x-\int \cos xe^{\tan x}\sec ^{2}xdx+\int \sec xe^{\tan x}dx$

$=\displaystyle \cos xe^{\tan x}+C$

$\displaystyle \int \left ( \log x \right )^{2}dx.$

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$\displaystyle x \left [\left ( \log x \right )^{2}-2 \log x+2 \right ].$
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$\displaystyle x \left [\left ( \log x \right )^{2}-2 \log x+1 \right ].$
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$\displaystyle x \left [\left ( \log x \right )^{2}-2 \log x-2 \right ].$
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$\displaystyle x \left [\left ( \log x \right )^{2}+2 \log x-2 \right ].$

Explanation

Integrating by parts,

$\displaystyle I=x \left ( \log x \right )^{2}-2\int x.\log x.\frac{1}{x}dx$

$\displaystyle =x \left ( \log x \right )^{2}-2\int \log x\:dx$

$\displaystyle =x \left ( \log x \right )^{2}-2\left [ x\log x-x \right ]$

$\displaystyle =x \left [\left ( \log x \right )^{2}-2 \log x+2 \right ].$

$\displaystyle \int x^{2}\tan ^{-1}x\:dx.$

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$\displaystyle \frac{x^{3}}{3}\tan^{-1}x-\frac{1}{6}x^{2}+\frac{1}{6}\log\left ( x^{2}+1 \right ).$
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$\displaystyle \frac{x^{3}}{3}\tan^{-1}x+\frac{1}{6}x^{2}+\frac{1}{6}\log\left ( x^{2}+1 \right ).$
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$\displaystyle \frac{x^{3}}{3}\tan^{-1}x-\frac{1}{3}x^{2}+\frac{1}{6}\log\left ( x^{2}+1 \right ).$
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$\displaystyle \frac{x^{3}}{3}\tan^{-1}x-\frac{1}{6}x^{2}+\frac{1}{3}\log\left ( x^{2}+1 \right ).$

Explanation

Integrating by parts,

$\displaystyle I=\frac{x^{3}}{3}\tan^{-1}x-\frac{1}{3}\int \frac{x^{3}}{x^{2}+1}dx.$

$=\displaystyle \frac{x^{3}}{3}\tan^{-1}x-\frac{1}{3}\int \left ( x-\frac{x}{x^{2}+1}\right )dx$

$=\displaystyle \frac{x^{3}}{3}\tan^{-1}x-\frac{1}{6}x^{2}+\frac{1}{6}\log\left ( x^{2}+1 \right )+c.$

$\displaystyle \int\frac{\log x}{(1+x)^{3}}dx$ is equal to

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$\displaystyle \frac{\log x}{(1+x)^{2}}+\displaystyle \frac{1}{2}\log\frac{x}{x+1}+\frac{1}{2}\frac{1}{x+1}+c$
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$\displaystyle 2\frac{-\log x}{(1+x)^{2}}+\displaystyle 2\log\frac{x+1}{x}+\frac{2}{x+1}+c$
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$\displaystyle \frac{-2\log x}{(1+x)^{2}}+\displaystyle 2\log\frac{x}{x+1}+\frac{2}{x+1}+c$
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$\displaystyle \frac{-\log x}{(1+x)^{2}}+\displaystyle \frac{1}{2}\log\frac{x+1}{x}-\frac{1}{2}\frac{1}{x+1}+c$

Explanation

$I=\int \dfrac { \log x }{ (1+x)^{ 3 } } dx$

$\displaystyle =\log x\int \dfrac { 1 }{ (1+x)^{ 3 } } dx-\int \dfrac { -2 }{ x(1+x)^{ 2 } } dx$

$\displaystyle =\dfrac { -2\log x }{ (1+x)^{ 2 } } +2\int \dfrac { 1 }{ x(1+x)^{ 2 } } dx$

Now,

$\displaystyle \dfrac { 1 }{ x(1+x)^{ 2 } } =\dfrac { A }{ x } +\dfrac { B }{ (1+x) } +\dfrac { C }{ (1+x)^{ 2 } }$

$\Rightarrow 1=A(1+x)^{ 2 }+Bx(1+x)+Cx$
When

$x=-1, C=-1$
When

$x=0,A=1$
When

$x=1,B=-1$
So

$\displaystyle \frac { 1 }{ x(1+x)^{ 2 } } =\frac { 1 }{ x } -\frac { 1 }{ (1+x) } -\frac { 1 }{ (1+x)^{ 2 } }$

$\displaystyle \int { \frac { 1 }{ x(1+x)^{ 2 } } dx } =\int { \frac { 1 }{ x } dx } -\int { \frac { dx }{ (1+x) } } -\int { \frac { dx }{ (1+x)^{ 2 } } }$

$\displaystyle =\log { x } - \log { (x+1) } +\frac { 1 }{ (x+1) }$

So,

$\displaystyle I=\frac { -2\log x }{ (1+x)^{ 2 } } +2\log { x } -2\log { (x+1) } +\frac { 2 }{ (x+1) } +C$

$\displaystyle I=\frac { -2\log x }{ (1+x)^{ 2 } } +2\log { \frac { x }{ x+1 } } +\frac { 2 }{ (x+1) } +C$

$\displaystyle \int x\sec^{2}2xdx$

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$\displaystyle \frac{1}{4}x\tan 2x-\frac{1}{2}\log \sec 2x.$
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$\displaystyle \frac{1}{2}x\tan 2x+\frac{1}{4}\log \sec 2x.$
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$\displaystyle \frac{1}{4}x\tan 2x-\frac{1}{4}\log \sec 2x.$
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$\displaystyle \frac{1}{2}x\tan 2x+\frac{1}{4}\log \cos 2x.$

Explanation

Let

$\displaystyle I=\int { x\sec ^{ 2 }{ 2x } dx }$

$\displaystyle =\frac { 1 }{ 2 } x\tan { 2x } -\frac { 1 }{ 2 } \int { \tan { 2x } dx }$

$\displaystyle =\frac { 1 }{ 2 } x\tan { 2x } -\frac { 1 }{ 4 } \log { \sec { 2x } }$

$\displaystyle =\frac { 1 }{ 2 } x\tan { 2x } +\frac { 1 }{ 4 } \log { \cos { 2x } }$
Hence, option 'D' is correct.

$\displaystyle\int \left ( \log x \right )^{2}dx.$

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$\displaystyle x\left ( \log x \right )^{2}+2x \log x+2x$
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$\displaystyle x\left ( \log x \right )^{2}-2x \log x+2x$
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$\displaystyle x\left ( \log x \right )^{2}-2x \log x$
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$\displaystyle x\left ( \log x \right )^{2}+2x \log x+x$

Explanation

$\displaystyle I= \int \left ( \log x \right )^{2}dx= \int 1.\left ( \log x \right )^{2}dx$

$\displaystyle = \left ( \log x \right )^{2}.x- \int 2\left ( \log x \right ).\frac{1}{x}.x\:dx$

$\displaystyle = x\left ( \log x \right )^{2}-2 \int 1.\log x dx$

$\displaystyle = x\left ( \log x \right )^{2}-2 \left [ x\log x-\int \frac{1}{x}.xdx \right ]$

$\displaystyle = x\left ( \log x \right )^{2}-2x \log x+2x$
Hence, option 'B' is correct.

$\displaystyle \int \left [ \left ( 1+x \right )e^{x}f\left ( x \right )+xe^{x}f'\left ( x \right ) \right ]dx=e^{x},$ then

$\displaystyle f\left ( x \right )=$

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$1$
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$x$
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$1/x$
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$\displaystyle e^{x}$

Explanation

$\displaystyle \frac{d}{dx}\left ( xe^{x} \right )=\left ( 1+x \right )e^{x}$

$\displaystyle \therefore I=\int f\left ( x\right )dx+ \int xe^{x}f'\left ( x \right )dx$
Integrating 1st by parts

$\displaystyle xe^{x}f\left ( x \right )-\int xe^{x}f'\left ( x \right )x+\int xe^{x}f'\left ( x \right )dx= xe^{x}f\left ( x \right )=e^{x}\therefore xf\left ( x \right )=1$
or

$\displaystyle f\left ( x \right )=1/x\Rightarrow \left ( c \right ).$

$\displaystyle \int \cos \sqrt{x}dx.$

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$\displaystyle \left [ \sqrt{x}\sin \sqrt{x}+\cos \sqrt{x} \right ].$
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$\displaystyle 2\left [ \sin \sqrt{x}-\cos \sqrt{x} \right ].$
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$\displaystyle 2\left [ \sqrt{x}\sin \sqrt{x}+\sqrt{x}\cos \sqrt{x} \right ].$
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$\displaystyle 2\left [ \sqrt{x}\sin \sqrt{x}+\cos \sqrt{x} \right ].$

Explanation

$\displaystyle I=\int \cos\sqrt{x}dx.$
Put

$\displaystyle \sqrt{x}=t$ so that

$\displaystyle \frac{1}{2\sqrt{\left ( x \right )}}dx=dt$

$\displaystyle \therefore I=f \left ( \cos t \right )2t\:dt=2\int t\cos t\:dt$
Integrating by parts, we get

$\displaystyle I=2\left [ t\sin t-\int 1.\sin t\:dt \right ]$

$\displaystyle =2\left [ t\sin t+\cos t \right ]$

$\displaystyle =2\left [ \sqrt{x}\sin \sqrt{x}+\cos \sqrt{x} \right ].$
Hence, option 'D' is correct.

$\displaystyle I=\int e^{x}\left [ \frac{1+\sqrt{1-x^{2}}\sin ^{-1}x}{\sqrt{\left ( 1-x^{2} \right )}} \right ]dx$

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$-e^{-x}\sin ^{-1}x+c$
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$-e^{x}\cos ^{-1}x+c$
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$e^{-x}\cos ^{-1}x+c$
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$e^{x}\sin ^{-1}x+c$

Explanation

$\displaystyle I=\int e^{x}\left [ \frac{1+\sqrt{1-x^{2}}\sin ^{-1}x}{\sqrt{\left ( 1-x^{2} \right )}} \right ]dx$

$\displaystyle =\int e^{x}.\frac{1}{\sqrt{\left ( 1-x^{2} \right )}}dx+\int e^{x}\sin^{-1}xdx.$
Integrate 1st by parts

$\displaystyle I=e^{x}.\sin^{-1}x-\int e^{x}\sin ^{-1}xdx+\int e^{x}\sin^{-1}xdx\:I=e^{x}\sin ^{-1}x.$ The last two integrals cancel.
Note : Above is also of the form

$\displaystyle \int e^{x}\left [ f\left ( x \right )+f'\left ( x \right )dx=e^{x}f\left ( x \right ) \right ]$

$\displaystyle f\left ( x \right )=\sin^{-1}x$ and

$\displaystyle f'(x)=\frac{1}{\sqrt{\left ( 1-x^{2} \right )}}.$

$\displaystyle\int \frac{x}{1+\cos x}dx.$

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$\displaystyle x\tan \left ( x/2 \right )-2\log \sec \left ( x/2 \right ).$
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$\displaystyle \tan \left ( x/2 \right )-2\log \sec \left ( x/2 \right ).$
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$\displaystyle x\tan \left ( x/2 \right )+2\log \sec \left ( x/2 \right ).$
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$\displaystyle x\tan \left ( x/2 \right )-\log \sec \left ( x/2 \right ).$

Explanation

$\displaystyle I=\int \frac{x}{1+\cos x}dx=\int \frac{x}{2\cos ^{2}\left ( x/2 \right )}dx$

$\displaystyle \int x\left [ \frac{1}{2}\sec^{2}\frac{x}{2} \right ]dx=x\tan\frac{x}{2}-\int 1.\tan \frac{x}{2}dx$

$\displaystyle =x\tan \left ( x/2 \right )-2\log \sec \left ( x/2 \right ).$
Hence, option 'A' is correct.

$\displaystyle\int \sin x\log \left ( \sec x+\tan x \right )dx.$

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$\displaystyle \cos x \log \left ( \sec x+\tan x \right )-x^2+c$
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$\displaystyle \sin x \log \left ( \sec x+\tan x \right )-x+c$
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$\displaystyle -\cos x \log \left ( \sec x+\tan x \right )-x+c$
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$\displaystyle -\sec x \log \left ( \sec x+\tan x \right )-x^2+c$

Explanation

$\displaystyle I= \int \sin x\log \left ( \sec x+\tan x \right )dx$

$\displaystyle I= \left ( -\cos x \right )\log \left ( \sec x+\tan x \right )-\int \left [ \frac{d}{dx}\log\left ( \sec x+\tan x \right ). \left ( -\cos x \right ) \right ]dx.$
Now

$\displaystyle \int \sec xdx=\log \left ( \sec x+\tan x\right );$

$\displaystyle \therefore \frac{d}{dx}\log \left (\sec x+\tan x \right )=\sec x$

$\displaystyle \therefore I=-\cos x \log \left ( \sec x+\tan x \right )-\int \sec x\left ( -\cos x \right )dx$

$\displaystyle =-\cos x \log \left ( \sec x+\tan x \right )-\int 1dx$

$\displaystyle =-\cos x \log \left ( \sec x+\tan x \right )-x +c$
Hence, option 'C' is correct.

Solve

$\displaystyle\int e^{x}\frac{1-\sin x}{1-\cos x}dx$

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$\displaystyle -e^{x}\cot \left (\dfrac {x}{2} \right )$
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$\displaystyle -e^{x}\tan \left (\dfrac{x} {2} \right )$
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$\displaystyle e^{x}\tan \left (\dfrac{x} {2} \right )$
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$\displaystyle e^{x}\cot \left (\dfrac {x}{2} \right )$

Explanation

$\displaystyle I=\int e^{x}\frac{\left ( 1-\sin x \right )}{\left ( 1-\cos x \right )}dx$

$\displaystyle I=\int \left [ \frac{e^{x}}{2\sin^{2}\left ( x/2 \right )}-e^{x}\frac{2\sin \left ( x/2 \right )\cos \left ( x/2 \right )}{2/sin^{2}\left ( x/2 \right )} \right ]dx$

$\displaystyle =\int e^{x}\left ( \frac{1}{2}co\sec^{2}\frac{x}{2} \right )dx-\int e^{x}\cot \frac{x}{2}dx.$
Integrate first by parts

$\displaystyle \therefore I=e^{x}\left ( -\cot \frac{x}{2} \right )-\int e^{x}\left ( -\cot \frac{x}{2} \right )dx-\int e^{x}\cot \frac{x}{2}dx$

$\displaystyle= -e^{x}\cot \left (\dfrac {x}{2} \right )$
The last two integrals cancel.

$\displaystyle \int 4^{x}\left [ g'\left ( x \right )+g\left ( x \right )\log 4\right ]dx=$

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$\displaystyle \frac{4^{x}}{\log 4}g\left ( x \right )$
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$\displaystyle 4^{x}$
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$\displaystyle 4^{x}\log 4.g\left ( x \right )$
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$\displaystyle 4^{x}g\left ( x \right )$

Explanation

$\displaystyle \int 4^{x}\left [ g'\left ( x \right )+g\left ( x \right )\log 4\right ]dx$

$\displaystyle I=\int 4^{x} g'\left ( x \right )dx+\int g\left ( x \right)4^{x}\log 4dx$

$=4^{ x }g(x)-\int { (4^{ x } } \log 4\int { g'\left( x \right) } dx)dx+\int g\left( x \right) 4^{ x }\log 4dx$

$=4^{ x }g(x)-\int { (4^{ x } } \log 4g(x)dx+\int g\left( x \right) 4^{ x }\log 4dx$

$\displaystyle \therefore I=4^{x}g\left ( x \right )+C$

$\displaystyle \int \frac{x}{1+\sec x}dx.$

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$\displaystyle \frac{x^{2}}{2}+x\tan \frac{x}{2}+2\log \sec \frac{x}{2}$
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$\displaystyle \frac{x^{2}}{2}-x\tan \frac{x}{2}+2\log \sec \frac{x}{2}$
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$\displaystyle \frac{x^{2}}{2}-\tan \frac{x}{2}+2\log \sec \frac{x}{2}$
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$\displaystyle \frac{x^{2}}{2}-x\tan \frac{x}{2}-2\log \sec \frac{x}{2}$

Explanation

$\displaystyle I=\int \frac{x}{\sec x+1}dx= \int \frac{x\cos x}{1+\cos x}dx$

$\displaystyle =\int x\frac{1+\cos x-1}{1+\cos x}dx= \int \left ( x-\frac{x}{2\cos^{2}\left ( x/2 \right )} \right )dx$

$\displaystyle I=\frac{x^{2}}{2}-x\tan \frac{x}{2}+2\log \sec \frac{x}{2}$

$\displaystyle \int \frac{co\sec^{2}x-203}{\left ( \cos x \right )^{203}}dx$

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$\displaystyle \frac{-\cot x}{\left ( \cos x \right )^{203}}$
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$\displaystyle\frac{\cos x}{\left ( co\sec x \right )^{203}}$
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$\displaystyle\frac{-\tan x}{\left ( co\sec x \right )^{203}}$
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$\displaystyle\frac{\cot x}{\left ( \sin x \right )^{203}}$

Explanation

$\displaystyle \int \left ( \cos^{-203} x\right )co\sec^{2}xdx-\int \frac{203}{\left ( \cos x \right )^{203}}dx$

$\displaystyle =I_{1}-I_{2}$
Integrate

$\displaystyle I_{1}$ by parts

$\displaystyle \therefore I_{1}=\left ( \cos ^{-203}x \right )\left ( -\cot x \right )-\int \cot x.203\cos ^{-204}x\left ( -\sin x \right )dx$

$\displaystyle I_{1}=-\frac{\cot x}{\left ( \cos x \right )^{203}}+\int \frac{203}{\left ( \cos x \right )^{203}}dx +C$

$\displaystyle \therefore I_{1}-I_{2}=-\frac{\cot x}{\left ( \cos x \right )^{203}}+C$

$\displaystyle \int a^{x}\left [ \log x+\log a\log \left ( \frac{x^{x}}{e^{x}} \right ) \right ]dx$

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$\displaystyle x\left ( \log x-1 \right )$
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$\displaystyle a^{x}.x\left ( \log x-1 \right )$
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$\displaystyle a^{x}.x\left ( \log x+1 \right )$
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$\displaystyle a^{x}.\left ( \log x-1 \right )$

Explanation

$\displaystyle I= \int a^{x}\log x dx+\int \left ( a^{x}\log a \right ).x\left ( \log x-1 \right )dx$
We know that

$\displaystyle \int \log x dx= x\log x -\int x.\frac{1}{x}dx=x\left ( \log x-1 \right )$
Integrating 1st term by parts,

$\displaystyle \therefore I=a^{x}.x\left ( \log x-1 \right )-\int x\left ( \log x-1 \right )\left ( a^{x}\log a \right )dx+\int \left ( a^{x}.\log a \right ).x\left ( \log x-1 \right )dx$

$\displaystyle =a^{x}.x\left ( \log x-1 \right )$

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 4 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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