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CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 2 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Indefinite Integrals
Quiz 2
$$ \int \sin ^{-1}(\cos x) d x $$ is equal to
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$$ \frac{\pi x}{2}+c $$
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$$ \frac{\pi x^{2}}{2}+c $$
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$$ \frac{\pi x-x^{2}}{2}+c $$
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$$ \frac{\pi x+x^{2}}{2}+c $$
Explanation
$$\text{sin}^{-1}(\cos x)=\dfrac{\pi}{2}-\text{cos}^{-1}(\cos x)=\dfrac{\pi }{2}- x$$
$$\displaystyle\int \text{sin}^{-1}(\cos x) d x=\int \bigg(\dfrac{\pi}{2}-x\bigg) d x=\dfrac{\pi x-x^2}{2}+C$$
What is $$\displaystyle \int \dfrac{dx}{x(1 + ln x)^n}$$ equal to $$(n \neq 1)$$ ?
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$$\dfrac{1}{(n - 1)(1 + ln x)^{n - 1}} + c$$
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$$\dfrac{1 - n}{(1 + ln x)^{1- n}} + c$$
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$$\dfrac{n + 1}{(1 + ln x)^{n+1}} + c$$
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$$-\dfrac{1}{(n - 1)(1 + ln x)^{n-1}} + c$$
Explanation
Given,
$$\int \dfrac{1}{x\left(1+\ln \left(x\right)\right)^n}dx$$
apply $$u=1+\ln \left(x\right)$$
$$=\int \dfrac{1}{u^n}du$$
$$=\int \:u^{-n}du$$
$$=\dfrac{u^{-n+1}}{-n+1}$$
$$=\dfrac{\left(1+\ln \left(x\right)\right)^{-n+1}}{-n+1}$$
$$=\dfrac{\left(1+\ln \left(x\right)\right)^{-n+1}}{-n+1}+C$$
$$-\dfrac{1}{(n - 1)(1 + ln x)^{n-1}} + c$$
Evaluate $$\displaystyle \int xsecx.tanxdx=$$
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$$x\sec x+\log|\tan(\pi/2+x/2)|+c$$
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$$x\sec x-\log $$$$|\tan(\pi/4+x/2)|+c$$
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$$x\sec x-\log |\tan(\pi/4+x)|+c$$
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$$ x\sec x+\log|\tan x/2|+c$$
Explanation
$$\int x\ \sec x\ \tan x\ dx=\ \int\ x(\sec x(\tan x))dx$$
$$=x\int(\sec x\ \tan x)dx$$ $$-\int1\int(\sec x\ \tan x)dx$$ ....... [Using integration parts]
$$=x \sec x-\int \sec x\ dx$$
$$=x\sec x-\log\left | \sec x+\tan \ x \right |+c$$
$$=x\sec x-\log\left | \tan (\pi/4+x/2) \right |+c$$ .....
$$[\because \sec x+\tan x=\tan (\pi/4+x/2)]$$
$$\therefore \int x\sec x \tan x\ dx=x\sec x-\log\left | \tan (\pi/4+x/2) \right |+c$$
$$\displaystyle \int\{\frac{(\log x-1)}{1+(\log x)^{2}}\}^{2}dx$$ is equal to
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$$\displaystyle \frac{x}{(\log x)^{2}+1}+c $$
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$$\displaystyle \frac{xe^{x}}{1+x^{2}}+c$$
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$$^{\dfrac{x}{x^{2}+1}+c} $$
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$$\displaystyle \frac{\log x}{(\log x)^{2}+1}+c$$
Explanation
$$\int [\dfrac {log x-1}{1+(log x)^2}]^2dx$$
$$\int \dfrac {(log x-1)^2}{[1+(log x)^2]^2}=\int \dfrac {(log x)^2+1-2 log x)}{[1+(log x)^2]^2}$$
$$=\int (\dfrac {1}{[(1+(log x)^2]}-\dfrac {2 log x}{[1+log^2x]^2})dx$$
$$=\int \dfrac {1}{(1+log^2x)}dx-\int \dfrac {2 log x}{(1+log^2x)^2}dx$$
$$=\dfrac {1}{(1+log^2x)}\int 1 dx+\int [\dfrac {2 log x}{(1+log^2x)^2}\times\dfrac {1}{x}\int 1\cdot dx]dx-\int \dfrac {2 log x}{(1+log^2x)^2}dx$$
$$=\dfrac {x}{1+log^2x}+\int \dfrac {2 log x}{(1+log^2x)^2}dx-\int \dfrac {2 log x}{(1+log^2x)^2}dx+c$$
$$=\dfrac {x}{1+log^2x}+c$$
$$\displaystyle \int \sec^{2}x.\text{cosec}^{2}xdx=$$
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$$ \tan x-\cot x +c$$
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$$\tan x +\cot x +c$$
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$$-\tan x + \cot x +c$$
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$$\sec x \tan x +c$$
Explanation
$$\displaystyle \int \sec^{2}x. \text{cosec}^{2}xdx$$
$$\displaystyle =\int \dfrac{1}{\cos^{2}x.\sin^{2}x}dx$$
$$\displaystyle =\int \dfrac{\cos^{2}x+\sin^{2}x}{\cos^{2}x.\sin^{2}x}dx$$
$$\displaystyle =\int \dfrac{\cos^{2}x}{\cos^{2}x\sin^{2}x}+\dfrac{\sin^{2}x}{\cos^{2}x.\sin^{2}x}dx$$
$$\displaystyle =\int \dfrac{1}{\sin^{2}x}+\dfrac{1}{\cos^{2}x}dx$$
$$\displaystyle =\int (\text{cosec}^{2}x+\sec^{2}x) dx$$
$$=-\cot x+\tan x +c$$
$$=\tan x-\cot x+c$$
$$\displaystyle \int\sqrt{\frac{x}{x-1}}dx, x\in(0,\pi/2)$$ equals
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$$\sqrt{x(x-1)}+\log(\sqrt{x}+\sqrt{x-1})+c$$
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$$\sqrt{x(x-1)}-\log(\sqrt{x}+\sqrt{x-1})+c$$
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$$\sqrt{x(x-1)}+\log(\sqrt{x}-\sqrt{x-1})+c$$
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$$\sqrt{x(x+1)}-\log(\sqrt{x}-\sqrt{x+1})+c$$
Explanation
$$x=sec^{2}\theta$$
$$dx =2 sec^{2}\theta \cdot tan \theta \cdot d\theta$$
$$\int \sqrt{\dfrac{sec^{2}\theta}{tan^{2}\theta}} \cdot2 sec^{2}\theta \ tan\ \theta d\theta$$
$$\int \dfrac{1}{sin \theta} \cdot 2 sec^{2}\theta \cdot \dfrac{sin \theta}{cos \theta} \cdot d\theta$$
$$=2\int sec^{3}\theta \cdot d\theta \cdot$$
$$=2\int (sec \theta)(sec^{2}\theta)d\theta \cdot$$
$$=2[sec\theta tan \theta -\int (sec \theta tan \theta \cdot tan \theta)d\theta \cdot ]$$
$$2\int sec^{3}\theta ^{d\theta}=2[sec\theta tan \theta]-2\int sec \theta(sec^{2}\theta-1)d\theta$$
$$2\int sec^{3}\theta d\theta =sec\theta tan \theta +\int sec \theta d\theta$$
$$2\int sec^{3} \theta d\theta = sec\theta tan\theta \rightarrow log |sec\theta+tan \theta |$$
and$$=\sqrt{x(x-1)}+log |\sqrt{x}+\sqrt{x-1}|+c$$
$$\int \dfrac{x}{\sqrt{x-1}} dx = \sqrt{x(x-1)} +log |\sqrt{x}+\sqrt{x-1}|+c \cdot$$
$$\displaystyle \int xe^{2x}(1+x)dx$$
equal to
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$$\displaystyle \frac{xe^{x}}{2}+c$$
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$$\displaystyle \frac{(e^{x})^{2}}{2}r$$
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$$\displaystyle \frac{(1+x)^{2}}{2}+c$$
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$$\displaystyle \frac{(xe^x)^2}{2}$$
Explanation
$$\displaystyle I=\int xe^{2x}(1+x)dx$$
$$\displaystyle =\int xe^{2x}dx+\int x^{2}e^{2x}dx$$
$$\displaystyle =e^{2x} \frac{x^{2}}{2}-\int x^{2}e^{2x}dx+\int x^{2}e^{2x}dx $$
$$\displaystyle I=\frac{(xe^{x})^{2}}{2}$$
The value of $$\displaystyle \int x^{2}2^{3x}dx=$$
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$$\displaystyle \frac{x^{2}2^{3x}}{4\log 2}-\frac{x.2^{3x+1}}{9.(\log 2)^3}+\frac{2^{3x+1}}{27(\log 2)}+c$$
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$$\displaystyle \frac{x^{2}2^{3x}}{3\log 2}-\frac{x.2^{3x+1}}{9.(\log 2)}+\frac{2^{3x+1}}{27(\log 2)}+c$$
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$$\displaystyle \frac{x^{2}2^{3x}}{3\log 2}-\frac{2^{3x+1}.x}{3^{2}(\log 2)^{2}}+\frac{2^{3x+1}}{3^{3}(\log 2)^{3}}+c$$
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$$\displaystyle \frac{x^{2}2^{3x}}{\log 2}-\frac{2^{3x+1}.x}{3^{2}(\log 2)^{2}}+\frac{2^{3x+1}}{(\log 2)^{3}}+c$$
Explanation
$$\int x^2\cdot 2^{3x} dx\cdot$$
$$=x^2\int 2^{3x}dx+\int 2x\int 2^{3x}\cdot dx\cdot$$
$$=x^2\dfrac {2^{3x}}{log 8}-\int 2x\dfrac {2^{3x}}{log 8^{dx}}$$
$$=\dfrac {x^22^{3x}}{log 8}-\dfrac {2}{log 8}[\dfrac {x2^{3x.}}{log 8}-\int \dfrac {2^{3x}}{log 8^{.dx}}]$$
$$=\dfrac {x^22^{3x}}{log 8}-\dfrac {2}{log 8}[\dfrac {x2^{3x}}{log 8}-\dfrac {1}{log 8}\dfrac {2^{3x}}{log 8}]+c.$$
$$=\dfrac {x^{2}2^{3x}}{log 8}-\dfrac {2^{3x+1}\cdot x}{3^2(log 2)^2}+\dfrac {2^{3x+1}}{3^3(log 2)^{3}}+c$$
$$\displaystyle \int\log{x}d{x}=$$
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$$x(\log x-x)+c$$
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$$ x(\log x)+c$$
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$$ x(\log x-1)+c$$
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$$ x(\log x+x)+c$$
Explanation
$$\int\ log\ x\ dx= \int\ log\ x-1\ dx$$
$$= log\ x\int\ 1-dx-\ \int(^1/_x.\sqrt{1.dx})dx$$
$$= log\ x(x)-x+c$$
$$= x(log\ x-1)+c$$
$$\because \int \ log\ x\ dx= x(log\ x-1)+c$$
Evaluate : $$\displaystyle \int\frac{\cot^{2}x}{(co\sec^{2}x+co\sec x)}d{x}$$
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$$ x-\sin x +C$$
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$$x + \cos x +C$$
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$$ \sin x - x+C $$
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$$2\tan(\displaystyle \frac{ax}{2})+c$$
Explanation
$$\displaystyle \int\frac{\cot^{2}x}{(co\sec^{2}x+co\sec x)}d{x}$$
$$\displaystyle \int \dfrac{cos ec^{2}x-1}{cos ecx(cosecx+1)}dx$$
$$\displaystyle \int \dfrac{(cosecx-1)(cosecx+1)}{cos ecx(cosecx+1)}dx$$
$$\displaystyle \int \dfrac{cosecx-1}{cosecx}dx=\int (1-sin x)dx$$
$$=x+cos x+c$$
$$\because -\int sin x.dx=cos x+c$$
$$\displaystyle \int[\frac{cosx}{x}-\sin{x}\log x]dx=$$
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$$ (\log x) sinx+c$$
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$$(\log x) (cos x) +c$$
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$$2(\log x) (cos x) +c$$
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$$-(\log x) sin x+c$$
Explanation
$$\int\left [ \displaystyle \dfrac{cos\ x}{x}-sinx\ log\ x \right ]dx$$
$$\int \displaystyle \dfrac{cos\ x}{x}dx-\ \int\ sin\ x\ log\ x\ dx$$
$$cos\ x\ \int \displaystyle \dfrac{1}{x}\ dx+\ \int\ sin\ x\ log\ x.\ dx$$ $$-\int\ sin\ x\ log\ x\ dx.$$
$$= (\ cos\ x)( log\ x)+c$$
Hence, $$\int \left [ \displaystyle \dfrac{cos x}{x}-(sin\ x)(log\ x) \right ]dx=(\ cos\ x)(\ log\ x)+c$$
Evaluate $$\displaystyle \int x^{2}e^{x}dx=$$
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$$ e^{x}(x^{2}-2x+2)+c$$
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$$ e^{x}(x^{2}+2x+2)+c$$
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$$ x^{2}+ex+c$$
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$$ e^{x}(x^{2}+x+2)+c$$
Explanation
Let $$I=\displaystyle \int x^{2}e^{x} dx $$
$$= x^{2} \displaystyle \int e^{x}dx$$ $$-\displaystyle \int \left (\frac{dx^{2}}{dx}\cdot \int e^{x} dx\right).dx$$
$$=x^{2} e^{x} -\displaystyle \int (2xe^{x}) dx\cdot$$
$$=x^{2}e^{x}-2\displaystyle \int xe^{x}\cdot dx\cdot$$
Taking $$\displaystyle \int xe^{x}= xe^{x}-e^{x}\cdot$$ ....(by using by parts)
Therefore, $$I=x^{2}e^{x}-2[xe^{x}-e^{x}]+c$$
$$=e^{x}[x^{2}-2x+2]+c$$
$$\displaystyle \int e^{\log x}\cos xdx=$$
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$$ xsinx-cosx+c$$
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$$\dfrac{x}{2}sinx+cos^{2}x+c$$
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$$ xsinx+cosx+c$$
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$$ xsinx+sinx+c$$
Explanation
$$\int e^{logx}cosx\ dx=\int x\ cosx\ dx.$$
$$=x\int cosx\ dx-\ \int \left ( 1\int cosxdx \right )dx$$
$$=x\ sinx+cosx+c$$
$$\int e^{logx}cosx\ dx=\ x\ sinx+cosx+c.$$
$$\displaystyle \int\frac{x\cos x\log x-\sin x}{x(\log x)^{2}}dx$$ is equal to
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$$\displaystyle \frac{\sin x}{\log x}+c$$
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$$ \log{x}+\sin{x+c}$$
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$$ \log{x}\sin{x+c}$$
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$$\displaystyle \frac{\cos x}{\log x}+c$$
Explanation
$$\displaystyle\int\frac{x\cos x\log\ x-\sin x}{x(\log\ x)^{2}}dx$$
$$=\displaystyle\int \frac{\cos x}{\log x} dx - \int\frac{\sin x}{x(\log\ x)^{2}}dx$$
Using Integration by parts in the first expression, we get
$$=\displaystyle\frac{1}{\log\ x}\int\cos x dx -\int\left(\frac{d\left (\dfrac{1}{\log x}\right )}{dx}\cdot\int\cos x dx\right)dx-\int \frac{\sin x}{x(\log x)^{2}}dx$$
$$=\displaystyle\frac{1}{\log x}(\sin x)+\int\frac{1}{x(\log x)^{2}}\sin x dx-\int \frac{\sin x}{x(\log x)^{2}}dx$$
$$=\dfrac{\sin x}{\log x}+c$$
where, $$c$$ is a constant of integration.
The value of $$\int x(cosecx\ cotx)dx=$$ is
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$$ xcosec x-log |tanx/2|+c$$
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$$2-xcosec x+\log| tan \displaystyle \frac{x}{2}|+c$$
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$$ xcosec x-2 \log|tanx/2|+c$$
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$$ xcot x-log |tan{\displaystyle \frac{x}{2}}|+c$$
Explanation
$$\int x(cosecx\ cot)dx=x\int (cosecx\ cotx)dx$$$$-\int 1\left ( \int cosecx\ cot\ xdx \right )dx$$
$$=x\left [ -cosecx \right ]+c+\int (cosecx)dx$$
$$=\ -x\ cosecx+c+log\left | cosecx-cotx \right |$$
$$=-x\ cosecx+c+\ log\left | \dfrac{1-cosx}{sinx} \right |$$
$$=2-x\ cosecx+c_{1}+log\left | tan\ x/2 \right |$$
$$\int x\ cosecx\ cotx\ dx$$
$$=2-x\ cosecx+c_{1}+log\ \left | tan\ \dfrac{x}{2} \right |$$
Solve $$\displaystyle \int \tan^{-1}xdx$$
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$$ x\tan x -\log|1+x^{2}|+c$$
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$$x$$ $$\tan^{-1} x -\displaystyle \frac{1}{2}\log|1+x^{2}|+c$$
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$$ x$$ $$\tan^{-1} x+\log\sqrt{1+x^{2}}+c$$
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$$ x\tan x+\log|1+x^{2}|+c$$
Explanation
$$\displaystyle \int \tan^{-1}x \ .1 \ \ dx$$
Let us assume here $$\tan^{-1}x$$ is the first function and constant 1 is the second function. Then the integral of the second function is x.
$$\displaystyle \int u.v \ dx=u\int v\ dx$$ $$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$$
Now by integrating the functions by parts, we get
$$\displaystyle =\tan^{-1}x\int 1\ dx$$ $$\displaystyle -\int \left ( d\frac{\ tan^{-1}x}{dx}\int 1\ dx \right )dx$$
$$\displaystyle =x\tan^{-1}x-\int \frac{1}{1+x^{2}} x\ dx$$
$$\displaystyle =x\tan^{-1}x+c-\frac{1}{2}\int \frac{2x\ dx}{1+x^{2}}$$
$$\displaystyle =-\ x\tan^{-1}x-\frac{1}{2}\log(1+x^{2})+c$$
$$=\displaystyle \int \tan^{-1}x\ dx=x\tan^{-1}x-\frac{1}{2}\log(1+x^{2})+c$$
Evaluate $$\displaystyle \int\frac{\log(x/e)}{(\log x)^{2}}dx$$
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$$\displaystyle \frac{\log x}{x}+c$$
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$$ \displaystyle \frac{x}{\log x}+c$$
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$$^{\dfrac{x}{log(x)^{2}}+c} $$
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$$\displaystyle \frac{(\log x)^{2}}{x}+c$$
Explanation
$$\displaystyle \int \dfrac{\log (x/e)}{(\log x)^{2}}dx=\int\ \dfrac{\log \ x-1}{(\log x)^{2}}$$
$$=\displaystyle \int\ \left [ \dfrac{1}{(\log \ x)}-\dfrac{1}{(\log x)^{2}} \right ]dx$$
$$=\displaystyle \int\ \dfrac{1}{(\log \ x)}dx-\ \int\ \dfrac{1}{(\log \ x)^{2}}dx$$
$$=\displaystyle \dfrac{1}{(\log \ x)}\int\ 1.dx+\ \int\ \dfrac{1}{(\log \ x)^{2}}dx-\int\ \dfrac{1}{(\log \ x)^{2}}dx$$ ..... [Using integration by parts]
$$=\ \dfrac{x}{\log \ x}+c$$
$$\therefore \displaystyle \int\ \dfrac{\log {(x/e)}}{(\log \ x)^{2}}dx=\dfrac{x}{\log x}+c$$
$$\displaystyle \int x^{2} {\it cosh} 4xdx=$$
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$$\displaystyle \frac{x^{2}}{4} {\it sinh}4x-\displaystyle \frac{x}{8}{\it cosh} 4x+\displaystyle \frac{\sinh 4x}{32}+c$$
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$$\displaystyle \frac{x^{2}}{4}sinh4x+\frac{x}{8}$$ cosh $$4{x-}\displaystyle \frac{\sinh 4x}{32}+c$$
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$$\displaystyle \frac{x^{2}}{4}sinh4x-\frac{x}{8}$$ cosh $$4x-\displaystyle \frac{\sinh 4x}{32}+c $$
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$$\displaystyle \frac{x^{2}}{4}sinh4x+\frac{x}{8}$$ cosh $$4x-\displaystyle \frac{\sin h4x}{64}+$$
Explanation
$$\int x^{2}\ cosh\ 4x.dx$$
$$x^{2}\int cosh\ 4x\ dx-\int 2x\left [ \left ( \int cosh\ 4x \right )dx \right ]dx$$
$$=\dfrac{x^{2}}{4}sin\ h4x-\ 1/2\ \int (x\ sinh4x)dx\ (1)$$
$$\int x\ sinh\ 4x\ dx=\dfrac{x}{4}\ cos\ h4x-\int \dfrac{cosh}{4}4x\ dx$$
$$=\left ( \dfrac{x}{4}cos\ h\ 4x-\dfrac{sin}{16} 4x\right )\ -(2)$$
Using (2) in (1),
$$=\dfrac{x^{2}}{4}sin\ h4x-\dfrac{x}{8}cosh\ 4x+\dfrac{sinh4x}{32}+c$$
$$\int x^{2}\ cosh\ 4x\ dx=\dfrac{x^{2}}{4}sin\ h4x-x/8\ cosh\ 4x+\dfrac{sinh\ 4x}{32}+c$$
$$\displaystyle \int cot^{-1}(\frac{x-1}{x+1})dx=$$
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$$ \displaystyle \frac{\pi}{4}x+x\cot^{-1}x-\frac{1}{2}\log(1+x^{2})+c$$
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$$ x\displaystyle \cot^{-1}x+\frac{1}{2}\log(1+x^{2})+c$$
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$$\displaystyle \frac{\pi}{4}x-\frac{1}{2}\log(1+x^{2})+c$$
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$$^{\displaystyle \frac{\pi}{4}x+x\cot^{-1}x}+\displaystyle \frac{1}{2}\log(1+x^{2})+c$$
Explanation
$$cosec\theta =\sqrt{1+x^{2}}$$
$$sin \theta =\dfrac{1}{\sqrt{1+x^{2}}}$$
$$\int cot^{-1}(\dfrac{x-1}{x+1})dx$$
$$x=cot\theta $$
$$\int cot^{-1}(\dfrac{cot\theta -1}{cot\theta +1})[cosec^{2}\theta ]d\theta $$
$$-\int cot^{-1}[cot(\pi /4+\theta )](cosec^{2}\theta )d\theta $$
$$=-\int (\pi /4+\theta (cosec^{2}\theta ))d\theta $$
$$=-[\int \pi /4cosec^{\gamma }\theta d\theta +\int \theta cosec^{\gamma \theta d\theta }]$$
$$=-[-\pi /4cot\theta +[\theta (-cot\theta )+\int cot\theta d\theta ]$$
$$=\pi /4cot\theta +cot\theta -log|sin\theta |+c$$
$$=\pi /4x+x cot^{-1}x-1/2log(\sqrt{1+x^{2}})+c$$
$$\int cot^{-1}(\dfrac{x-1}{x+1})dx=\pi /4{x}+x cot^{-1}x -1/2 log (\sqrt{1+x^{2}})+c$$
$$\displaystyle \int e^{x}cosec x(1-\cot x)dx=$$
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$$ e^{x}\cot x+c$$
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$$ e^{x}cosec x\cot x+c$$
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$$ e^{x}cosec x+c$$
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$$-e^{x}\cot x+c$$
Explanation
$$\int e^{x}cosec\ x(1-cot\ x)dx$$
$$\int e^{x}\left ( cosec\ x-cose\ x\ cot\ x \right )dx$$
$$\int e^{x}cosec\ x\ dx-\int e^{x}\left ( cosec\ x.cot\ x \right )dx$$
$$\int e^{x}\left ( cosec\ x\ cot\ x \right )dx=ex\int cosec\ x\ cot\ x\ dx+\int e^{x}cosec\ x.dx$$
$$=e^{x}cosec\ x+\int cosec\ x\ dx.$$
$$\int e^{x}cosec\ x\ dx-\left ( -e^{x}cosec\ x+\int cosec\ x\ dx \right )+c$$
$$=\int e^{x}cosec\ x\ dx+e^{x}\ cosec\ x-\int e^{x}\ cosec\ x\ dx$$
$$=\ e^{x}cosec\ x+c$$
$$\displaystyle \int x^{2}a^{x}dx=$$
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$$ a^{x}[\displaystyle \frac{x^{2}}{\log a}-\frac{2x}{(\log a)^{2}}+\frac{2}{(\log a)^{3}}]+c$$
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$$ a^{x}[\displaystyle \frac{x^{2}}{\log a}+\frac{2x}{(\log a)^{2}}-\frac{2}{(\log a)^{3}}]+c$$
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$$ a^{x}[\displaystyle \frac{x^{2}}{\log a}+\frac{2x}{(\log a)^{2}}+\frac{2}{(\log a)^{3}}]+c$$
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$$ a^{x}(\displaystyle \frac{x}{\log a}-\frac{1}{(\log a)^{2}})+c$$
Explanation
$$\int x^{2}a^{x}dx = x^{2}\int a^{x}dx-\int \left [ \dfrac{dx^{2}}{dx}\int a^{x}\ dx \right ]dx$$
$$\dfrac{x^{2}a^{x}}{log\ a}-\int \left [ 2x\int a^{x}dx \right ]dx$$
$$\dfrac{x^{2}a^{x}}{log\ a}-2\int \dfrac{x\ a^{x}}{log^{a}}dx$$
$$\dfrac{x^{2}a^{x}}{log\ a}-\dfrac{2}{log^{a}}\int [xa^{x}]dx$$
$$\int xa^{x}dx=\dfrac{xa^{a}}{ ^{a}}-\int 1\dfrac{a^{x}}{(log^{a})}dx$$
$$=\dfrac{xa^{x}}{log^{a}}-\dfrac{a^{x}}{(log^{a})^{2}}$$
$$=\dfrac{x^{2}a^{x}}{log^{a}}-\dfrac{2}{log^{a}}\left [ \dfrac{xa^{x}}{log^{a}}-\dfrac{a^{x}}{(log^{a})^{2}} \right ]$$
$$=\dfrac{x^{2}a^{x}}{log^{a}}-\dfrac{2}{log^{a}}\left [ \dfrac{xa^{x}}{log^{a}}-\dfrac{a^{x}}{(log^{a})^{2}} \right ]$$
$$a^{x}\left [ \dfrac{x^{2}}{log{a}}-\dfrac{2x}{(loga)^{2}}+\dfrac{2}{(log\ a)^{3}} \right ]+c$$
lf polynomials $$P$$ and $$Q$$ satisfy $$\displaystyle{\int((3x-1)\cos x+(1-2x)\sin x)dv}=P\cos x+Q\sin x+R$$ (ignoring the constant of integration) then
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$$ P=3x-2$$
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$$ Q=2+x$$
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$$ P=3(x-1)$$
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$$Q=3(x-1)$$
Explanation
Let $$I=\int((3x-1)\cos x+(1-2x)\sin x)dx$$
Using integration by parts, we evaluate the 2 expressions individually,
$$ \int (3x-1) \cos x = (3x - 1) \sin x - \int 3\sin x dx $$
$$= (3x - 1) \sin x + 3 \cos x + c $$
Similarly,
$$ \int (1-2x)\sin x dx = -(1-2x)\cos x - \int (-2)(-\cos x) dx $$
$$= (2x - 1) \cos x - 2\sin x + c $$
Adding both the expressions, we get
$$I= 3(x - 1)\sin x + 2(x+1)\cos x $$ Ignoring the constant
Comparing it with the RHS, we get
$$ P = 2(x+1)$$ and $$Q = 3(x-1)$$
Hence, option 'D' is correct.
Evaluate $$\displaystyle \int\frac{\arcsin\sqrt{x}}{\sqrt{1-x}}dx$$ $$=$$
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$$2[\sqrt{x}-\sqrt{1-x}$$ arc $$\sin\sqrt{x}]+c$$
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$$2[\sqrt{x}+\sqrt{1-x}$$ arc $$\sin\sqrt{x}]+c$$
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$$2[\sqrt{x}+\sqrt{1-x}$$ arc $$\cos\sqrt{x}]+c$$
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$$2[\sqrt{x}-\sqrt{1-x}$$ arc $$\cos\sqrt{x}]+c$$
Explanation
$$\displaystyle \int \dfrac {\sin ^{-1}\sqrt x}{\sqrt {1-x}}\cdot dx.$$
Put $$x=\sin ^2\theta$$
$$\Rightarrow dx=2 \sin \theta \cos \theta d\theta.$$
$$\therefore \displaystyle \int \dfrac {\sin ^{-1}\sqrt x}{\sqrt {1-x}}\cdot dx.$$
$$=\displaystyle \int \dfrac {\sin ^{-1}(\sin \theta)}{\sqrt {1-\sin ^2\theta}}\cdot d\theta(2 \sin \theta)(\cos \theta)$$
$$\displaystyle =2\int \theta \sin \theta d\theta$$
$$=2[\theta(-\cos \theta)+\int \cos \theta d\theta.]$$
$$=2[\theta(-\cos \theta)+\sin \theta]+c$$
$$=2[\sin \theta-\theta \cos \theta]+c$$
$$=2[\sqrt x-(\sin ^{-1}\sqrt x)\sqrt {1-x}]+c$$
$$\therefore \displaystyle \int \dfrac {\sin ^{-1}\sqrt x}{\sqrt {1-x}}dx=2[\sqrt x-(\sqrt {1-x})\sin ^{-1}\sqrt x]+c$$
$$\displaystyle \int e^{x}(x^{2}-5x+8)$$ dx $$=e^{x}f(x)+c$$ then $$f(x)$$
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$$x^{2}-5x+12$$
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$$ x^{2}+7x+15$$
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$$ x^{2}-7x-15$$
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$$ x^{2}-7x+15$$
Explanation
$$\int e^{x}(x^{2}-5x+8)dx$$
$$=\int e^{x}x^{2}dx-5\int e^{x}.xdx+8\int e^{x}\ dx$$
$$=x^{2}e^{x}=2\int xe^{x}\ dx-5\int e^{x}\ x+8\int e^{x}dx$$
$$x^{2}e^{x}-7[xe^{x}-e^{x}]+8e^{x}$$
$$=x^{2}e^{x}-7xe^{x}+15\ e\ ^{x}$$
$$=e^{n}[x^{2}-7x+15]+c$$
$$so,\ f(x)-\ x^{2}-7x+15$$
Evaluate $$\displaystyle \int x\frac{(\sec 2x-1)}{(\sec 2x+1)}dx$$
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$$x \tan x-\log |\displaystyle \sec x|+\frac{x^{2}}{2}+c$$
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$$x \tan x-\log |\displaystyle \sec x|-\frac{x^{2}}{2}+c$$
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$$x \tan x-\log |\sec\frac{x}{2}|+c$$
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$$x \tan x-\log |\displaystyle \sec\frac{x}{2}|+\frac{x^{2}}{2}+c$$
Explanation
$$\displaystyle \int x\left(\dfrac{\sec 2x-1}{\sec 2x+1}\right)dx=\int x\left(\dfrac{1-\cos 2x}{1+\cos 2x}\right)dx$$
$$=\displaystyle \int x \tan^{2}x dx$$
$$\displaystyle \int x(\sec ^{2}x-1)dx$$
$$\displaystyle =\int x \sec ^{2}x \ {dx}-\int xdx$$
We know, $$\displaystyle \int u.v \ dx=u\int v\ dx$$ $$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$$
Now by integrating the functions by parts, for the first part taking $$x$$ as $$u $$ and $$sec^2 x$$ as $$v$$.
$$\displaystyle =x\int \sec ^{2}xdx-\int 1\left(\int \sec ^{2}xdx\right)dx-\int xdx$$
$$\displaystyle =x \tan x-\int \tan x \ dx-\dfrac{x^{2}}{2}+c$$
$$=x \tan x-\log |\sec x|-\dfrac{x^{2}}{2}+c$$
$$\displaystyle \int x\left(\dfrac{\sec 2x-1}{\sec 2x+1}\right)dx=x \tan x-\log|\sec x|-\dfrac{x^{2}}{2}+c$$
$$\displaystyle \int e^{x}2^{3\log_{2}x}dx=e^{x}f(x)+c$$, then $$f(x)=$$
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$${ x }^{ 3 }-3{ x }^{ 2 }+6x-6$$
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$$x^{3}-3x^{2}-6x-3$$
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$$ x^{3}-3x^{2}+6x+6$$
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$$x^{3}+3x^{2}+6x+6$$
Explanation
Given, $$\displaystyle \int e^{x}2^{3\log_{2}x}dx=e^{x}f(x)+c$$ ......(A)
Consider, $$\int e^{ x }2^{ 3\log _{ 2 }{ x } }dx=\int e^{ x }.x^{ 3 }dx$$
$$\int e^{x} \times x^{3}dx=x^{3}\int e^{x}dx-\int 3x^{2}e^x\ dx$$
$$=x^{3}e^{x}-3 \int x^{2}e^{x}\ dx\ $$ ....(1)
$$\int x^{2}e^{x}=x^{2}e^{x}-2\int xe^{x}\ dx\ $$ .....(2)
$$\int xe^{x}=\ xe^{x}-e^{x}\ $$ .....(3)
Using (3) in (2),
$$\int x^{2}e^{x}=x^{2}e^{x}-2(xe^{x}-e^{x})\ $$ ....(4)
Using (4) in (1)
$$ \displaystyle \int e^{ x }.x^{ 3 }dx= x^{3}e^{x}-3\left [ x^{2}e^{x}-2(xe^{x}-e^{x}) \right ]+c$$
$$=x^{3}e^{x}-3x^{2}e^{x}+6xe^{x}-6e^{x}+c$$
$$=e^{x}\left [ x^{3}-3x^{2}+6x-6 \right ]+c$$
On comparing with eqn (A), we get
$$f(x)=x^{3}-3x^{2}+6x-6$$
$$\displaystyle \int\frac{x+\sin x}{1+\cos x}dx=$$
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$$xtan \displaystyle \frac{x}{2}+c$$
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$$xcot \displaystyle \frac{x}{2}+c$$
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$$xsin \displaystyle \frac{x}{2}+c$$
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$$xcos \displaystyle \frac{x}{2}+c$$
Explanation
$$\displaystyle \int\frac{x+\sin x}{1+\cos x}dx$$
$$\displaystyle =\frac { 1 }{ 2 } \int \frac { x+\sin x }{ \cos ^{ 2 }{ (x/2 )} } dx$$
$$\displaystyle =\frac { 1 }{ 2 } \int x\sec ^{ 2 }{ (x/2) } dx+\int \tan { (x/2) } dx$$
$$\displaystyle=\frac { 1 }{ 2 } [x\frac { \tan { (x/2) } }{ 1/2 } -\int \frac { \tan { (x/2) } }{ 1/2 } dx]+\int \tan { (x/2) } dx+C$$
$$=x\tan { (x/2) } +C$$
$$\displaystyle \int\cos^{-1}(\frac{1}{x})dx$$ equal to
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$$ x\sec^{-1}x+\cosh^{-1}x+c$$
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$$ x\sec^{-1}x-\cosh^{-1}x+c$$
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$$ x\sec^{-1}x-\sin^{-1}x+c$$
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$$ x\sec^{-1}x+\sin^{-1}x+c$$
Explanation
$$x=sec\theta$$
$$dx=sec\theta \cdot tan\theta d\theta.$$
$$\int cox^{-1}(cos\theta)\cdot (sec\theta \cdot tan\theta)d\theta.$$
$$\int \theta (sec\theta \cdot tan\theta)d\theta \cdot \theta \int (sec\theta \cdot tan\theta)d\theta -\int (\int(sec\theta \cdot tan\theta)d\theta)d\theta$$
$$\int sec\theta tan\theta\cdot d\theta=sec\theta+c$$
$$=\theta sec\theta-\int sec\theta d\theta$$
$$=\theta sec\theta+(log \mid sec\theta+tan\theta \mid)+1.$$
$$=x sec^{-1}x.-log \mid x+\sqrt {x^2-1}\mid +c$$
$$=x sec^{-1}x-cos h^{-1}x+c.$$
$$\int cos^{-1}(\dfrac {1}{x})=x sec^{-1}{x}-cos h^{-1}x+c.$$
Evaluate: $$\displaystyle \int\log_{10}xdx$$
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$$ (x-1)\log_{e}x+c$$
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$$ \displaystyle \log_{e}10.x\log_{e}(\frac{x}{e})+c$$
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$$ \displaystyle \log_{10}e.x\log_{e}(\frac{x}{e})+c$$
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$$\cfrac{1}{x}+c$$
Explanation
We know that,
$$\int { uvdx } =u'\int { vdx } -\int { u'(\int { vdx } )dx } +C$$ (Integration by parts rule)
$$\therefore$$ we can write,
$$I=\int { \log _{ 10 }{ x } dx } =\log _{ 10 }{ e } \int { \log _{ e }{ x } \cdot (1)dx }$$ ...as $$\log _{ 10 }{ x } =\cfrac { \log _{ e }{ x } }{ \log _{ e }{ 10 } } =\log _{ 10 }{ e } \cdot \log _{ e }{ x } $$
Applying integration by parts;
$$u=\log _{ e }{ x } v=1\\ u'=\cfrac { 1 }{ x } \int { vdx } =x\\ $$
$$\therefore I=\log _{ 10 }{ e } ((\log _{ e }{ x } )x-\int { \cfrac { 1 }{ x } \cdot xdx } )+C=\log _{ 10 }{ e } (x\log _{ e }{ x } -\int { dx } )+C\\ I=\log _{ 10 }{ e } \cdot x(\log _{ e }{ x } -1)+C=\log _{ 10 }{ e } \cdot x(\log _{ e }{ x } -\log _{ e }{ e } )+C\\ \therefore I=\log _{ 10 }{ e } \cdot x(\log _{ e }{ \cfrac { x }{ e } } )+C$$
Evaluate $$\displaystyle \int\frac{xe^{x}}{(x+1)^{2}}dx$$
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$$ \displaystyle e^{x}[\dfrac{1}{x+1}]+c$$
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$$ \displaystyle e^{x}[\dfrac{-1}{x+1}]+c$$
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$$ e^{x}x+c$$
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$$\displaystyle e^{x}[\dfrac{-1}{(x+1)^{2}}]+c$$
Explanation
$$I = \displaystyle \int\frac{xe^{x}}{(x+1)^{2}}dx$$
$$\displaystyle =\int\frac{(x+1-1)e^{x}}{(x+1)^{2}}dx$$
$$\displaystyle =\int \frac{(x+1)e^{x}}{(x+1)^{2}}dx -\int\frac{e^{x}}{(x+1)^{2}}dx$$
$$\displaystyle =\int \frac{e^{x}}{x+1} dx-\int\frac{e^{x}}{(x+1)^{2}}dx$$
We know, $$\displaystyle \int u.v \ dx=u\int v\ dx$$ $$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$$
Now by integrating the functions by parts, for the first part taking $$e^x$$ as $$u $$ and $$\dfrac{1}{x+1}$$ as $$v$$
$$I\displaystyle =\frac{e^{x}}{x+1} +\int\frac{e^{x}}{(x+1)^{2}}dx-\int\frac{e^{x}}{(x+1)^{2}}dx$$
$$\displaystyle I=\frac{e^{x}}{x+1}+C$$
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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