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CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 4 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Indefinite Integrals
Quiz 4
If $$\displaystyle \int x$$ 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.
If $$\displaystyle \int x^{2}e^{-2x}dx=e^{-2x}(ax^{2}+bx+c)+d$$ 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,
$$\displaystyle
\therefore I=a^{x}.x\left ( \log x-1 \right )-\int x\left ( \log x-1
\right )\left ( a^{x}\log a \right )dx+\int \left ( a^{x}.\log a \right
).x\left ( \log x-1 \right )dx$$
$$\displaystyle =a^{x}.x\left ( \log x-1 \right )$$
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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