MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 9

$\displaystyle\int \left\{\dfrac{(log x-1)}{1+(log x)^2}\right\}^2dx$ is equals to?

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$\dfrac{log x}{(log x)^2+1}+C$
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$\dfrac{x}{x^2+1}+C$
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$\dfrac{xe^x}{1+x^2}+C$
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$\dfrac{x}{(log x)^2+1}+C$

Explanation

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

let logx = t

$\Rightarrow dx = e^{t}.dt (since x = e^{t})$

$I = \displaystyle\int \left(\dfrac{logx-1}{1+(logx)^{2}}\right)^{2}dx$

$= \displaystyle\int \left(\dfrac{t-1}{1+t^{2}}\right)^{2}e^{t}.dt$

$= \displaystyle\int \dfrac{t^{2}+1-2t}{(1+t^{2})^2}e^{t}.dt$

$= \displaystyle\int e^{t}\left(\dfrac{t^{2}+1}{(t^{2}+1)^{2}}-\dfrac{2t}{(1+t^{2})^{2}}\right).dt$

$= \displaystyle\int e^{t}\left(\dfrac{1}{t^{2}+1}-\dfrac{2t}{(1+t^{2})^{2}}\right)dt$

$= e^{t}(\dfrac{1}{t^{2}+1})+c$

$= e^{logx}\left(\dfrac{1}{(logx)^{2}+1}\right)+c$

$= \dfrac{x}{(logx)^{2}+1}+c$

The value of the integral

$\int _ { - \pi } ^ { \pi } ( \cos p x - \sin q x ) ^ { 2 } d x$ where

$p , q$ are integers, is equal to:-

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$- \pi$
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0
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$\pi$
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2p

$\int \frac { \cos x + 2 \sin x } { 7 \sin x - 5 \cos x } d x = a x + b \ln | 7 \sin x - 5 \cos x | + c$ then

$a+b$ is

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$\frac { 1 } { 17 }$
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$\frac { 3 } { 37 }$
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$\frac { 13 } { 37 }$
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$\frac { 23 } { 47 }$

Explanation

$\displaystyle\int \frac{\cos x+2 \sin x}{7 \sin x- 5 \cos x}$

Let

$\displaystyle 2 \sin (x)+ \cos (x)= \frac{9}{74}(7 \sin x- 5 \cos x)+\frac{17}{74}(5 \sin x+ 7 \cos x)$

$\displaystyle \Rightarrow \int \frac{9/74(7 \sin x- 5 \cos x)+17/74(5 \sin x+7 \cos x)}{(7 \sin x- 5 \cos x)}dx$

$\displaystyle \Rightarrow \int \dfrac{9}{74}dx+\frac{17}{74}\int \frac{5 \sin x+7 \cos x}{7 \sin x-5 \cos x}dx$

Let

$u= 7 sin x- 5 cos x$

$\therefore dx=\dfrac{1}{5\sin x+7\cos x}du$

$\displaystyle\therefore \frac{9}{74}x+\dfrac{17}{74}\int \dfrac{1}{u}du$

$\dfrac{9}{74}x+\dfrac{17}{74}\log(u)$

$\therefore \dfrac{9}{74}x+\dfrac{17}{74}\log(7 \sin x-5 \cos x)+c$

$\therefore a=\dfrac{9}{74}, b=\dfrac{17}{74}$

$\therefore a+b\Rightarrow \dfrac{26}{74}\Rightarrow \dfrac{13}{37}$

$\displaystyle \int \dfrac { d x } { \sin ^ { 2 } x \cos ^ { 2 } x }$ equals-

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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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$\tan x - \cot 2 x + c$

Explanation

$I=\int \dfrac{dx}{\sin^{2}x\,\cos^{2}x}$

$I= \int \dfrac{\sin^{2}x+cos^{2}x}{\sin^{2}x\,cos^{2}x}dx$

$I= \int (\dfrac{1}{cos^{2}x}+\dfrac{1}{sin^{2}x})dx = \int (\sec^{2}x+cosec^{2}x)dx$

$I=\ tanx-\cot x+c$

$\displaystyle \int \frac { 1 - x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) \sqrt { 1 + x ^ { 4 } } } d x$ is equal to

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$\sqrt { 2 } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$
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$\frac { 1 } { \sqrt { 2 } } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$
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$\frac { 1 } { 2 } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$
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$\frac { 1 } { \sqrt { 2 } } \sin ^ { - 1 } \left\{ \frac { x ^ { 2 } + 1 } { \sqrt { 2 } x } \right\} + c$

$\displaystyle\int { \dfrac { \left( x+3 \right) { e }^{ x } }{ { \left( x+4 \right) }^{ 2 } } } dx$ is equal to

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$\dfrac { 1 }{ { \left( x+4 \right) }^{ 2 } } +C$
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$\dfrac { { e }^{ x } }{ { \left( x+4 \right) }^{ 2 } } +C$
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$\dfrac { { e }^{ x } }{ x+4 } +C$
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$\dfrac { { e }^{ x } }{ x+3 } +C$

Explanation

$I=\displaystyle\int { \dfrac { (x+3){ e }^{ x } }{ { (x+4) }^{ 2 } } dx }$

$=\displaystyle\int { \dfrac { (x+4-1){ e }^{ x } }{ { (x+4) }^{ 2 } } dx }$

$=\displaystyle\int { \dfrac { { e }^{ x } }{ { (x+4) } } dx }$

$-\displaystyle\int { \dfrac { { e }^{ x } }{ { (x+4) }^{ 2 } } dx }$

$=\dfrac{e^x}{x+4}+\displaystyle\int { \dfrac { { e }^{ x } }{ { (x+4)^2 } } dx }$

$-\displaystyle\int { \dfrac { { e }^{ x } }{ { (x+4) }^{ 2 } } dx }$ [ Using method of by parts]

$=\dfrac{e^x}{x+4}+c$ [ Where

$c$ is integrating constant].

$\int \frac { x ^ { 2 } \tan ^ { - 1 } x } { 1 + x ^ { 2 } } d x = \tan ^ { - 1 } x - \frac { 1 } { 2 } \log \left( 1 + x ^ { 2 } \right) + f ( x ) + c$ then

$f ( x ) =$

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$- \frac { \tan ^ { - 1 } x } { 2 }$
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$- \frac { 1 } { 2 } \left( \tan ^ { - 1 } x \right) ^ { 2 }$
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$\frac { \tan ^ { - 1 } x } { 2 }$
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None of these

Explanation

$\\\int\>(\frac{(1+x^2-1)tan^{-1}x}{1+x^2})dx\\=\int\>\left(1-(\frac{1}{1+x^2})\right)tan^{-1}x\>dx\\=\int\>1.tan^{-1}dx-\int\>(\frac{tan^{-1}x}{1+x^2})dx\\Use\>ILATE\>in\>first\>part\>and\>assume\\tan^{-1}x=t\>for\>second\>integration\\=xtan^{-1}x-\int\>(\frac{1}{1+x^2})\times\>xdx-\int\>t\>dt\\=xtan^{-1}x-(\frac{1}{2})log|1+x^2|-(\frac{t^2}{2})+C\\=xtan^{-1}x-(\frac{1}{2})log|1+x^2|-(\frac{(tan^{-1}x)^2}{2})+C\\\therefore\>C=-(\frac{1}{2})(tan^{-1}x)^2$

$\int _{ 0 }^{ 1 }{ \dfrac { dx }{ \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+2 \right) } }$ =

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$\dfrac { \pi }{ 4 } +\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } }$
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$\dfrac { \pi }{ 2 } -\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } }$
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$\dfrac { \pi }{ 4 } -\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } }$
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$\dfrac { \pi }{ 3 } -\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } }$

The value of

$\int _{ -1 }^{ 1 }{ \dfrac { { cot }^{ -1 }x }{ \pi } } dx$

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1
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2
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3
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0

$\int \left( x ^ { 6 } + 7 x ^ { 5 } + 6 x ^ { 4 } + 5 x ^ { 3 } + 4 x ^ { 2 } + 3 x + 1 \right) e ^ { x } d x$ equals

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$\sum _ { j = 0 } ^ { 6 } x ^ { j } e ^ { x } + c$
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$\sum _ { j = 1 } ^ { 7 } x ^ { j } e ^ { x } + c$
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$\sum _ { i = 1 } ^ { 6 } x ^ { i } e ^ { x } + c$
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$\sum _ { i = 0 } ^ { 5 } x ^ { i } e ^ { x } + c$

Explanation

$\displaystyle\int (x^6+7x^5+6x^4+5x^3+4x^2+3x+1)e^xdx$

$=\displaystyle\int (x^6+6x^5)e^xdx+\displaystyle\int (x^5+5x^4)e^xdx+\displaystyle\int (x^4+4x^3)e^xdx+\displaystyle\int (x^3+3x^2)e^xdx=\displaystyle\int (x^2+2x)e^xdx+\displaystyle\int (x+1)e^xdx$ .

$\displaystyle\int [f(x)+f'(x)]e^xdx=f(x)e^x$

Using this fact, we get

$\displaystyle\int (x^6+6x^5)e^xdx=\displaystyle\int (x^6+(x^6)')e^xdx$

$=x^6\cdot e^xdx$

Hence G.E

$=x^6e^x+x^5e^x+...…+xe^x+c$

$=\displaystyle\sum^6_{i=1}x^ie^x+c$ .

1186161_1246215_ans_d3782b273b9942879d93c6f4d7ba7563.jpg

The integral

$\int _{ 2a/4 }^{ a/2 }{ (2\quad cosecx{ ) }^{ 17 } }$ dx is equal to:

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$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ 2({ e }^{ u }+{ e }^{ -u }{ ) }^{ 16 } } du$
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$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ ({ e }^{ u }+{ e }^{ -u }{ ) }^{ 17 } } du$
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$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ ({ e }^{ u }-{ e }^{ -u }{ ) }^{ 17 } } du$
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$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ 2({ e }^{ u }-{ e }^{ -u }{ ) }^{ 16 } } du$

Evaluate:

$\displaystyle\int { \dfrac { { x }^{ 2 }+1 }{ { x }^{ 4 }+1 } dx }$ equals

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$\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{x^{2}-1}{\sqrt{2}x}\right)+C$
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$\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{1-x^{2}}{\sqrt{2}x}\right)+C$
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$\dfrac{1}{2}\tan^{-1}\left(\dfrac{x^{2}-1}{\sqrt{2}x}\right)+C$
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$\dfrac{1}{2}\tan^{-1}\left(\dfrac{1-x^{2}}{\sqrt{2}x}\right)+C$

Explanation

$\displaystyle \int \dfrac{x^2 + 1}{x^4 + 1} . dx$

$\displaystyle \int \dfrac{1 + x^{-2}}{x^2 + x^{-2}}. dx$ [divide with

$x^2$ ]

$\displaystyle \int \dfrac{1 + x^{-2}}{x^2 + x^{-2} -2 + 2} .dx$

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

$u = x - x^{-1}$

$\displaystyle \int \dfrac{dU}{U^2 + 2} = \dfrac{1}{\sqrt{2}} \tan^{-1} \dfrac{U}{\sqrt{2}} + C$

$= \dfrac{1}{\sqrt{2}} \tan^{-1} \dfrac{(x - x^{-1})}{\sqrt{2}} + C$

$= \dfrac{1}{\sqrt{2}} \tan^{-1} \left(\dfrac{x^2 - 1}{\sqrt{2} x} \right) + C$

Solve:

$\int {\dfrac{{dx}}{{\left( {x - 3} \right)\sqrt {x + 1} }}}$

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$\cos { { h }^{ -1 } } \left( \dfrac { 1-x }{ \sqrt { 3 } \left( 1+x \right) } \right) +c$
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$\sin { { h }^{ -1 } } \left( \dfrac { 1-x }{ \sqrt { 3 } \left( 1+x \right) } \right) +c$
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$-\sin { { h }^{ -1 } } \left( \dfrac { 1-x }{ \sqrt { 3 } \left( 1+x \right) } \right) +c$
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$-\cos { { h }^{ -1 } } \left( \dfrac { 1-x }{ \sqrt { 3 } \left( 1+x \right) } \right) +c$

$\displaystyle \int{\dfrac{1-x^{2}}{(1+x^{2})\sqrt{1+x^{4}}}dx}$ is equal to

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$\sqrt{2}\sin^{-1}\left\{\dfrac{\sqrt{2}x}{x^{2}+1}\right\}+c$
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$\dfrac{1}{\sqrt{2}}\sin^{-1}\left\{\dfrac{\sqrt{2}x}{x^{2}+1}\right\}+c$
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$\dfrac{1}{2}\sin^{-1}\left\{\dfrac{\sqrt{2}x}{x^{2}+1}\right\}+c$
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$\dfrac{1}{\sqrt{2}}\sin^{-1}\left\{\dfrac{x^{2}+1}{\sqrt{2}x}\right\}+c$

Explanation

$\begin{array}{l} \int { \frac { { 1+{ x^{ 2 } } } }{ { \left( { 1+{ x^{ 2 } } } \right) \sqrt { 1+{ x^{ 4 } } } } } dx } \\ \int { \frac { { \left( { 1-{ x^{ 2 } } } \right) } }{ { { x^{ 2 } }\left( { x+\frac { 1 }{ x } } \right) \sqrt { { x^{ 2 } }+\frac { 1 }{ { { x^{ 2 } } } } } } } dx } \\ =\int { \frac { { \left( { \frac { 1 }{ { { x^{ 2 } } } } -1 } \right) dx } }{ { \left( { x+\frac { 1 }{ x } } \right) \sqrt { { x^{ 2 } }+\frac { 1 }{ { { x^{ 2 } } } } } } } } \\ =-\int { \frac { { \left( { 1-\frac { 1 }{ { { x^{ 2 } } } } } \right) dx } }{ { \left( { x+\frac { 1 }{ x } } \right) \sqrt { { { \left( { x+\frac { 1 }{ x } } \right) }^{ 2 } }-2 } } } } \\ x+\frac { 1 }{ x } =t \\ \frac { { dt } }{ { dx } } =1-\frac { 1 }{ { { x^{ 2 } } } } \\ =-\int { \frac { { dt } }{ { t\sqrt { { t^{ 2 } }-1 } } } } \\ =-\frac { 1 }{ { \sqrt { 2 } } } { \sec ^{ -1 } }\left( { \frac { t }{ { \sqrt { 2 } } } } \right) +C \\ =-\frac { 1 }{ { \sqrt { 2 } } } \left[ { \frac { \pi }{ 2 } -\cos e{ c^{ -1 } }\frac { 1 }{ { \sqrt { 2 } } } } \right] +C \\ =\frac { 1 }{ { \sqrt { 2 } } } \cos e{ c^{ -1 } }\left( { \frac { 1 }{ { \sqrt { 2 } } } } \right) +C' \\ =\frac { 1 }{ { \sqrt { 2 } } } { \sin ^{ -1 } }\left( { \frac { { \sqrt { 2 } } }{ t } } \right) +C' \\ =\frac { 1 }{ { \sqrt { 2 } } } { \sin ^{ -1 } }\left( { \frac { { \sqrt { 2 } } }{ { x+\frac { 1 }{ x } } } } \right) +C' \\ =\frac { 1 }{ { \sqrt { 2 } } } { \sin ^{ -1 } }\left( { \frac { { \sqrt { 2 } x } }{ { { x^{ 2 } }+1 } } } \right) +C' \\ Hence,\, option\, B\, is\, the\, correct\, answer. \end{array}$

The value of the integral

$\int _{ 0 }^{ 1 }{ \sqrt { \frac { 1-x }{ 1+x } } }$ dx is

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$\frac { \pi }{ 2 } +1$
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$\frac { \pi }{ 2 } -1$
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-1
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1

$\displaystyle \int{\dfrac{x^{3}-1}{x^{3}+x}dx}$ equal to

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$x-\log x+\log(x^{2}+1)-\tan^{-1}x+c$
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$x-\log x+\dfrac{1}{2}\log(x^{2}+1)-\tan^{-1}x+c$
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$x+\log x+\dfrac{1}{2}\log(x^{2}+1)+\tan^{-1}x+c$
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$x+\log x-\dfrac{1}{2}\log(x^{2}+1)-\tan^{-1}x+c$

Explanation

$I=\int { \dfrac { { { x^{ 3 } }-1 } }{ { { x^{ 3 } }+x } } dx } \\ I=\int { \dfrac { { \left( { x-1 } \right) \left( { { x^{ 2 } }+x+1 } \right) } }{ { x\left( { { x^{ 2 } }+1 } \right) } } dx } \\ I=\int { \dfrac { { x-1 } }{ x } +\int { \dfrac { { x-1 } }{ { { x^{ 2 } }+1 } } dx } }$

$I =\int 1 dx-\int \dfrac{1}{x} dx+\int { \dfrac { { x-1 } }{ { { x^{ 2 } }+1 } } dx }$

$I=x-\ln { x } +\int { \dfrac { { x } }{ { { x^{ 2 } }+1 } } dx } -\int \dfrac{1}{x^2+1}dx$

$I =x-{ \log _{ e } }x+\dfrac { 1 }{ 2 } { \log _{ e } }\left( { { x^{ 2 } }+1 } \right) -{ \tan ^{ -1 } }x+C$

Hence, this is the answer.

The integral

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

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$(2x-1).e^{x^{2-\dfrac {1}{x}}}+c$
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$(2x+1).e^{x^{2-\dfrac {1}{x}}}+c$
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$xe^{x^{2-\dfrac {1}{x}}}+c$
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$-xe^{x^{2-\dfrac {1}{x}}}+c$

Explanation

$\begin{array}{l} Let \\ I=\int { { e^{ { x^{ 2 } }-\frac { 1 }{ x } } }\left( { 1+2{ x^{ 2 } }+\frac { 1 }{ x } } \right) dx } \\ =\int { { e^{ { x^{ 2 } }-\frac { 1 }{ x } } }\left( { x\left( { 2x+\frac { 1 }{ { { x^{ 2 } } } } } \right) +1 } \right) dx } \\ =\int { { e^{ g } }\left( { fg'+f' } \right) dx } \\ ={ e^{ g } }f+C \\ ={ e^{ x-\frac { 1 }{ { { x^{ 2 } } } } } }\times x+C \\ Hence,\, option\, C\, is\, the\, correct\, answer. \end{array}$

$\displaystyle \int e^{x/2}\sin (\dfrac {x}{2}+\dfrac {\pi}{4})dx$ is equal to---

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$e^{x/2}\sin x/2+c$
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$e^{x/2}\cos x/2+c$
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$\sqrt {2}e^{x/2}\sin x/2+c$
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$\sqrt {2}e^{x/2}\cos x/2+c$

Explanation

$\begin{array}{l} \int { { e^{ \frac { x }{ 2 } } }.\sin \left( { \frac { x }{ 2 } +\frac { \pi }{ 4 } } \right) dx } \\ let\, \, \left( { \frac { x }{ 2 } +\frac { \pi }{ 4 } } \right) =t \\ \frac { { dx } }{ 2 } =dt \\ dx=2dt \\ \Rightarrow 2\int { { e^{ t-\frac { \pi }{ 4 } } }\sin t } \\ \Rightarrow \frac { 2 }{ { { e^{ \frac { \pi }{ 4 } } } } } \int { { e^{ t } }\sin t=\sin t. } \int { { e^{ t } }dt } -\int { \left[ { \frac { { d\sin t } }{ { dt } } \int { { e^{ t } }dt } } \right] dt } \\ \Rightarrow \frac { 2 }{ { { e^{ \frac { \pi }{ 4 } } } } } \int { { e^{ t } }\sin t={ e^{ t } }\sin t+ } \int { \cos t.{ e^{ t } }\, dt } \\ \Rightarrow \frac { 2 }{ { { e^{ \frac { \pi }{ 4 } } } } } \int { { e^{ t } }\sin t } ={ e^{ t } }\sin t+\cos t\int { { e^{ t } }dt } -\int { \frac { { d\cos t } }{ { dt } } } \int { \left. { { e^{ t } }dt } \right] } dt \\ \Rightarrow \frac { 2 }{ { { e^{ \frac { \pi }{ 4 } } } } } \int { { e^{ t } }\sin t={ e^{ t } }\sin t+\cos t.{ e^{ t } } } -\int { \sin t.{ e^{ t } }dt } \\ \Rightarrow \int { { e^{ t } }\sin t\left[ { \frac { 2 }{ { { e^{ \frac { \pi }{ 4 } } } } } +1 } \right] ={ e^{ t } }\left( { \sin t+\cos t } \right) +c } \\ \Rightarrow \int { { e^{ t } }\sin t=\frac { { \sqrt { 2 } { e^{ t } }\sin \left( { t+\frac { \pi }{ 4 } } \right) \times { e^{ \frac { \pi }{ 4 } } } } }{ { \left( { 2+{ e^{ \frac { \pi }{ 4 } } } } \right) } } +c } \\ =\frac { { \sqrt { 2 } { e^{ \left( { \frac { x }{ 2 } =\frac { \pi }{ 2 } } \right) } }{ { \sin }^{ \left( { \frac { x }{ 2 } +\frac { \pi }{ 2 } } \right) } } } }{ { 2+{ e^{ \frac { \pi }{ 4 } } } } } . \end{array}$

$\displaystyle \int \frac { e ^ { x } ( 1 + \sin x ) } { 1 + \cos x } d x =$

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

Explanation

Let

$I=\displaystyle\int\dfrac{e^x(1+\sin x)}{1+\cos x}dx$

$=\displaystyle\int e^x\left(\dfrac{1+2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}{2\cos^2\dfrac{x}{2}}\right)$

$=\displaystyle\int e^x\left(\dfrac{\sin^2\dfrac{x}{2}+\cos^2\dfrac{x}{2}+2\sin\dfrac{x}{2}.\cos\dfrac{x}{2}}{2\cos^2\dfrac{x}{2}}\right)$

$=\displaystyle\int e^x\left(\dfrac{\left(\sin\dfrac{x}{2}+\cos\dfrac{x}{2}\right)^2}{2\cos^2\dfrac{x}{2}}\right)$

$=\displaystyle\int \dfrac{e^x}{2}\left(\dfrac{\sin\dfrac{x}{2}+\cos\dfrac{x}{2}}{\cos\dfrac{x}{2}}\right)^2$

$=\displaystyle\int\dfrac{e^x}{2}\left(\dfrac{\sin\dfrac{x}{2}}{\cos\dfrac{x}{2}}+\dfrac{\cos\dfrac{x}{2}}{\cos\dfrac{x}{2}}\right)^2$

$=\displaystyle\int\dfrac{e^x}{2}\left(\tan\dfrac{x}{2}+1\right)^2$

$=\displaystyle\int\dfrac{e^x}{2}\left(\tan^2\dfrac{x}{2}+1+2\tan\dfrac{x}{2}\right)$

$=\displaystyle\int\dfrac{e^x}{2}\left(\sec^2\dfrac{x}{2}+2\tan\dfrac{x}{2}\right)$

$=\displaystyle\int e^x\left(\dfrac{1}{2}.\sec^2\dfrac{x}{2}+\dfrac{2}{2}\tan\dfrac{x}{2}\right)$

$=\displaystyle\int e^x\left(\tan\dfrac{x}{2}+\dfrac{1}{2}\sec^2\dfrac{x}{2}\right)$

Above is in the form of

$e^x[f(x)+f'(x)]dx=e^xf(x)+c$

$=e^x\tan\dfrac{x}{2}+c$

$\therefore$

$\displaystyle\int\dfrac{e^x(1+\sin x)}{1+\cos x}dx$

$=e^x\tan\dfrac{x}{2}+c$

$\displaystyle \int \dfrac {1}{\sqrt {\sin^{3}x\sin(x+a)}}dx$ is equal to

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$2\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \tan x}+c$
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$-2\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \cot x}+c$
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$\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \tan x}+c$
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$-\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \tan x}+c$

Explanation

$\begin{array}{l} \int { \frac { { dx } }{ { \sqrt { { { \sin }^{ 3 } }x\sin \left( { x+a } \right) } } } } \\ =\int { \frac { { dx } }{ { \sqrt { { { \sin }^{ 3 } }x\sin x\left( { \cos a+\cot x\sin a } \right) } } } } \\ =\int { \frac { { \cos e{ c^{ 2 } }xdx } }{ { \sqrt { \cos a+\sin a\cot x } } } } \\ \cos a+\sin a\cot =t \\ \frac { { dt } }{ { dx } } =-\sin a\cos e{ c^{ 2 } }x \\ -\frac { 1 }{ { \sin a } } dt=\cos e{ c^{ 2 } }xdx \\ =-\frac { 1 }{ { \sin a } } \int { \frac { { dt } }{ { \sqrt { t } } } } \\ =-\frac { 1 }{ { \sin a } } \int { { t^{ \frac { { -1 } }{ 2 } } }dt } \\ =\frac { { -2 } }{ { \sin a } } { t^{ \frac { 1 }{ 2 } } }+C \\ =-2\cos eca\sqrt { \cos a+\sin a\cot x } +C \end{array}$

For

$x\in R,\ f(x)=|\log 2-\sin x|$ and

$g(x)=f(f(x))$ , then

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$g'(0)=\cos (\log 2)$
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$g'(0)=-\cos (\log 2)$
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$g$ is differentible at $x=0$ and $g'(0)=-\sin (\log 2)$
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$g$ is not differentiable at $x=0$

Explanation

In the neighourwood of

$x=0$

$f(x)=\log 2-\sin x$

$g(x)=f(x)=\log 2-\sin f(x)$

$=\log 2-\sin (\log 2-\sin x)$

$g'(x)=\cos (\log2-\sin x)(-\cos x)$

$g'(0)=\cos (log 2)$

option

$A$ is correct answer.

Solve:

$\displaystyle \int e^x \left(\tan^{-1}x + \dfrac{1}{1 + x^2}\right) dx$

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$e^x \tan^{-1} x + c$
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$\dfrac{e^x}{1 + x^2} + c$
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$e^x \tan x + c$
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None of these

Explanation

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

$\displaystyle\int e^x(f(x)+f'(x))dx=e^xf(x)$

$\therefore I=e^x\tan^{-1}x+c$ .

1126241_1289846_ans_7d3afa8313e54b65a33c892fb1deff51.JPG

$\int\dfrac{1}{\sqrt{x}+\sqrt{x+1}}dx$ is equal to

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$\dfrac{2}{3}[(x+1)^{3/2}-x^{3/2}]+C$
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$\dfrac{2x}{3}[\sqrt{x+1}-\sqrt{x}]+C$
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$\ln[\sqrt{x}-\sqrt{x+1}]+C$
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$\ln[\sqrt{x+1}-\sqrt{x}]+C$

Explanation

$\begin{array}{l} I=\int { \dfrac { 1 }{ { \sqrt { x } +\sqrt { x+1 } } } dx } \\I= \int { \dfrac { 1 }{ { \sqrt { x } +\sqrt { x+1 } } } \times \dfrac { { \sqrt { x } -\sqrt { x+1 } } }{ { \sqrt { x } -\sqrt { x+1 } } } dx } \\I= \int { \dfrac { { \sqrt { x } -\sqrt { x+1 } } }{ { x-x-1 } } dx } \\ I=-1\left[ { \int { \sqrt { x } dx } I=-\int { \sqrt { x+1 } dx } } \right] \\ I=-\left[ { \dfrac { { { x^{ 3/2 } } } }{ { \dfrac { 3 }{ 2 } } } -\dfrac { { { { \left( { x+1 } \right) }^{ 3/2 } } } }{ { \dfrac { 3 }{ 2 } } } } \right] +c \\ I=\dfrac { 2 }{ 3 } \left[ { { { \left( { x+1 } \right) }^{ 3/2 } }-{ x^{ 3/2 } } } \right] +c \end{array}$

Hence, this is the answer.

$\displaystyle\int \dfrac { x - 2 } { x ^ { 2 } - 4 x + 3 } d x =$

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$\log \left| \sqrt { x ^ { 2 } - 4 x + 3 } \right| + c$
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$\log | ( x - 3 ) ( x - 1 ) / + c$
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$x \log | x - 3 | - 2 \log | x - 2 | + x$
0%
None

Explanation

$\displaystyle\int \dfrac{x-2}{x^2+4x+3}dx$

$=\dfrac{1}{2}\displaystyle\int \dfrac{2x-4}{x^2-4x+3}dx=\dfrac{1}{2}\displaystyle\int \dfrac{d(x^2-4x+3)}{x^2-4x+3}$

$=\dfrac{1}{2}log|x^2-4x+3|+c$

$=log|\sqrt{x^2-4x+3}|+c$ .

The value of the definite integral

$\overset { { a }_{ 1 } }{ \underset { { a }_{ 2 } }{ \int { \frac { d\theta }{ 1+tan\theta } } } } =\frac { 501\pi }{ K }$ where

$\ a _{ 2 }=\quad \frac { 1003\pi }{ 2008 }$ and

${ \ a }_{ 1 }=\frac { \pi }{ 2008 }$ The value of K equalls

0%
2007
0%
2006
0%
2009
0%
2008

Explanation

$\begin{array}{l} \int _{ { a_{ 2 } } }^{ { a_{ 1 } } }{ \frac { { d\theta } }{ { 1+\tan \theta } } } \\ I=\int _{ \frac { \pi }{ { 2008 } } }^{ \frac { { 1003\pi } }{ { 2008 } } }{ \frac { { d\theta } }{ { 1+\tan \theta } } } \\ I=\int _{ \frac { \pi }{ { 2008 } } }^{ \frac { { 1003\pi } }{ { 2008 } } }{ \frac { { \tan \theta } }{ { 1+\tan \theta } } }d\theta \\ 2I=\int _{ \frac { \pi }{ { 2008 } } }^{ \frac { { 1003\pi } }{ { 2008 } } }{ d\theta } \\ 2I=\frac { { 1002\pi } }{ { 2008 } } \\ I=\frac { { 501 } }{ { 2008 } } \pi \\ k=2008 \end{array}$

Option

$D$ is correct answer.

$g\left( x \right) =\int { { x }^{ x }\log _{ e }{ (ex)dx } }$ then

$g\left( \pi \right)$ equals

0%
$\pi \log _{ e }{ \pi }$
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${ \pi }^{ \pi }\log _{ e }{ (e\pi } )$
0%
${ \pi }^{ \pi }\log _{ e }{ (\pi } )$
0%
${\pi}^\pi$

Explanation

$\begin{array}{l} g\left( x \right) =\int _{ }^{ }{ { x^{ x } } } \left( { 1+\log { e^{ x } } } \right) dx \\ =\int _{ }^{ }{ d\left( { { x^{ x } } } \right) } \\ g\left( x \right) ={ x^{ x } } \\ g\left( \pi \right) ={ \pi ^{ \pi } } \end{array}$

$\int {\dfrac{1}{{9{x^2} - 25}}dx = \_\_\_\_\_\_ + c.}$

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$\dfrac{1}{{30}}\log \left| {\dfrac{{3x + 5}}{{3x - 5}}} \right|$
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$\log \left| {x + \sqrt {3x - 5} } \right|$
0%
$\dfrac{1}{{30}}\log \left| {\dfrac{{3x - 5}}{{3x + 5}}} \right|$
0%
$\log \left| {x - \sqrt {3x - 5} } \right|$

Explanation

$I=\int { \dfrac { 1 }{ {9{ x^{ 2 } }-25 } } dx }$

$I=\dfrac { 1 }{ 9 } \int { \dfrac { 1 }{ { \left( { { x^{ 2 } }-\dfrac { { 25 } }{ 9 } } \right) } } dx }$

$I=\dfrac { 1 }{ 9 } \int { \dfrac { 1 }{ { { { x^{ 2 } }-\left(\dfrac { 5 }{ 3 }\right)^2 } } } dx }$

We know that

$\int \dfrac { dx }{ x^ 2-a^2}=\dfrac{1}{2a}\log\left(\dfrac{x-a}{x+a}\right)+C$

Therefore,

$I =\dfrac { 1 }{ 9 } .\dfrac { 1 }{ { 2\times \dfrac { 5 }{ 3 } } } \log \left| { \dfrac { { x-\dfrac { 5 }{ 3 } } }{ { x+\dfrac { 5 }{ 3 } } } } \right| +C$

$I=\dfrac { 3 }{ { 90 } } \log \left| { \dfrac { { \dfrac { { 3x-5 } }{ 3 } } }{ { \dfrac { { 3x+5 } }{ 3 } } } } \right| +C$

$I =\dfrac { 1 }{ { 30 } } \log \left| { \dfrac { { 3x-5 } }{ { 3x+5 } } } \right| +C$

Hence, this is the answer.

$I=\displaystyle \int _{ 0 }^{ 1 }{ \frac { dx }{ \left( 1+x \right) \left( 2+x \right) \sqrt { x\left( 1-x \right) } } }$ then

$I=$

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$2\pi$
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$\pi$
0%
$\dfrac {\pi}{2}$
0%
$\dfrac {\pi}{\sqrt {6}}|\sqrt {3}-1|$

$\int {{e^{3{{\log }_e}x}}.{{\left( {{x^4} + 1} \right)}^{ - 1}}dx = \_\_\_\_\_\_\_\_\_ + C.}$

0%
$\log \left( {{x^4} + 1} \right)$
0%
$\frac{1}{4}\log \left( {{x^4} + 1} \right)$
0%
$-\log \left( {{x^4} + 1} \right)$
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$\frac{{ - 3}}{{{{\left( {{x^4} + 1} \right)}^3}}}$

Explanation

$I=\int { { e^{ 3\log_ex } }.{ { \left( { { x^{ 4 } }+1 } \right) }^{ -1 } }dx }$

$I=\int { { e^{ \log_ex^3 } }.{ { \left( { { x^{ 4 } }+1 } \right) }^{ -1 } }dx }$

$I=\int \dfrac{ { x^3 }}{{ { \left( { { x^{ 4 } }+1 } \right) } }}dx$

Let

$t=x^4+1$

$dt=4x^3dx$

Therefore,

$I=\dfrac{1}{4}\int \dfrac{1}{t}dt$

$I=\dfrac{1}{4}\log(t)+C$

Put the value of

$t$ , we get

$I=\dfrac{1}{4}\log(x^4+1)+C$

Hence, this is the answer.

The integral of

$\dfrac{x^{2}-x}{x^{3}-x^{2}+x-1}$ w.r.t

$x$ is

0%
$\dfrac{1}{2}\log|x^{2}+1|+C$
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$\dfrac{1}{2}\log|x^{2}-1|+C$
0%
$\log|x^{2}+1|+C$
0%
$\log|x^{2}-1|+C$

Explanation

$\begin{array}{l} \int { \dfrac { { { x^{ 2 } }-x } }{ { { x^{ 3 } }-{ x^{ 2 } }+x-1 } } dx } \\ \Rightarrow \int { \dfrac { { x\left( { x-1 } \right) } }{ { { x^{ 2 } }\left( { x-1 } \right) +1\left( { x-1 } \right) } } dx } \\ \Rightarrow \int { \dfrac { x }{ { \left( { { x^{ 2 } }+1 } \right) } } dx } \\ Let\, \, { x^{ 2 } }+1=t \\ \Rightarrow 2x\, \, dx=dt \\ \Rightarrow \dfrac { 1 }{ 2 } \int { \dfrac { { dt } }{ t } =\dfrac { 1 }{ 2 } \log t+c\, \, \, \, =\dfrac { 1 }{ 2 } \log \left( { { x^{ 2 } }+1 } \right) +c\, \, \, \, \, \, Ans. } \end{array}$

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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