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CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 10 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Indefinite Integrals
Quiz 10
$$\int {{{(\sin x)}^{99}}{{(\cos x)}^{ - 101}}dx = \_\_\_\_\_\_\_ + C.} $$
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$$\dfrac{{{{(\tan x)}^{100}}}}{{100}}$$
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$$\dfrac{{{{(\tan x)}^2}}}{2}$$
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$$\dfrac{{{{(\tan x)}^{98}}}}{{98}}$$
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$$\dfrac{{{{(\tan x)}^{97}}}}{{97}}$$
Explanation
$$\begin{array}{l}I= { \int { { { \left( { \sin x } \right) }^{ 99 } }.\left( { \cos x } \right) } ^{ -101 } }dx. & \\I= \int { \dfrac { { { { \left( { \sin x } \right) }^{ 99 } } } }{ { { { \left( { \cos x } \right) }^{ 101 } } } } dx } & \\I= \int { { { \left( { \dfrac { { \sin x } }{ { \cos x } } } \right) }^{ 99 } }.\dfrac { 1 }{ { { { \cos }^{ 2 } }x } } dx } & \\I= \int { { { \left( { \tan x } \right) }^{ 99 } }.{ { \sec }^{ 2 } }xdx } & Put\, \, \tan x=t.\, \, { \sec ^{ 2 } }xdx=dt \\I= \int { { t^{ 99 } }dt } & \\I= \dfrac { { { t^{ 100 } } } }{ { 100 } } +c & \\ I=\dfrac { { { { \left( { \tan x } \right) }^{ 100 } } } }{ { 100 } } +c & \end{array}$$
Hence, this is the answer.
Evaluate $$\int {{x^2}\log x\,dx\, }$$.
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$$\cfrac{{{x^2}}}{2}\log x - \cfrac{1}{{9}}{x^2}+c$$
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$$\cfrac{{{x^3}}}{3}\log x - \cfrac{1}{{9}}{x^2}+c$$
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$$\cfrac{{{x^3}}}{3}\log x - \cfrac{1}{9}{x^3}+c$$
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$$\cfrac{{{x^3}}}{3}\log x + \cfrac{1}{9}{x^3}+c$$
Explanation
We have,
$$I=\displaystyle \int {x^2}\log x\,dx$$
We have,
$$\displaystyle \int u.v \ dx=u\int v\ dx$$ $$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$$
$$\displaystyle I=\log x\int x^2\ dx -\int \left(\dfrac{d}{dx}\log x \int x^2\ dx\right)\ dx$$
$$\displaystyle I=\log x\dfrac{x^3}{3} -\int \left(\dfrac{1}{x} \dfrac{x^3}{3}\right)\ dx$$
$$\displaystyle I=\dfrac{x^3\log x}{3} -\int \left( \dfrac{x^2}{3}\right)\ dx$$
$$\displaystyle I=\dfrac{x^3\log x}{3} - \dfrac{x^3}{9}+C$$
Hence, this is the answer.
The value of $$\int { { e }^{ x } } .\dfrac { { x }^{ 2 }+1 }{ { \left( x+1 \right) }^{ 2 } } dx$$ is
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$${ e }^{ x }\left( \dfrac { x-1 }{ x+1 } \right) +C$$
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$${ e }^{ x }\left( \dfrac { x+1 }{ x-1 } \right) +C$$
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$${ e }^{ x }.x+C$$
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None of these
Explanation
$$I=\displaystyle\int e^x\dfrac{x^2+1}{(x+1)^2}dx$$
$$=\displaystyle\int e^x\dfrac{x^2+1+2x-2x}{(x+1)^2}dx$$
$$=\displaystyle\int e^x\dfrac{(x+1)^2-2x}{(x+1)^2}dx$$
$$=\displaystyle\int e^x\left\{1-\dfrac{2x}{(x+1)^2}\right\}dx$$
$$=\displaystyle\int e^xdx-\displaystyle\int e^x\cdot \dfrac{2x}{(x+1)^2}dx$$.
$$=e^x-I_1+e$$
Let $$I_1=\displaystyle\int e^x\cdot \dfrac{2x}{(x+1)^2}dx$$
Here $$\dfrac{2x}{(x+1)^2}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+1)^2}$$
$$2x=A(x+1)+B$$
If $$x=-1$$,
$$2(-1)=B$$
$$\Rightarrow B=-2$$
If $$x=1$$,
$$2=2A+(-2)$$
$$2A=2+2$$
$$A=\dfrac{4}{2}=2$$
$$\therefore \dfrac{2x}{(x+1)^2}=\dfrac{2}{x+1}-\dfrac{2}{(x+1)^2}$$
$$\therefore I_1=\displaystyle\int e^x\left\{\dfrac{2}{x+1}-\dfrac{2}{(x+1)^2}\right\}dx$$
$$=\displaystyle\int 2e^x\left\{\dfrac{1}{x+1}-\dfrac{1}{(x+1)^2}\right\}dx$$
$$=2\displaystyle\int e^x\left\{\dfrac{1}{x+1}-\dfrac{1}{(x+1)^2}\right\}dx$$
Which is of the form $$\displaystyle\int e^x[f(x)+f'(x)]dx$$
$$=2e^x\cdot \left(\dfrac{1}{x+1}\right)$$ $$[\because\displaystyle\int e^x[f(x)+f'(x)]dx=e^xf(x)+c]$$
$$=\dfrac{2e^x}{x+1}$$
$$\therefore I=e^x-\dfrac{2e^x}{x+1}+c$$
$$=e^x\left(1-\dfrac{2}{x+1}\right)+c$$
$$=e^x\left\{\dfrac{x+1-2}{x+1}\right\}+c$$
$$\therefore I=e^x\left(\dfrac{x-1}{x+1}\right)+c$$.
$$\int {{e^x}(\log \sin x + \cot x)\,dx = } \_\_\_\_\_\_ + C.$$
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$${e^x}\cot x$$
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$${e^x}\log \sin x$$
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$${e^x}\tan x$$
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None of these
Explanation
$$\int {{e^x}\left( {\log \sin x + \cot x} \right)dx} $$
As we know
$$\begin{array}{l} \int { { e^{ x } }\left[ { f\left( x \right) +f'\left( x \right) } \right] dx } ={ e^{ x } }f\left( x \right) +c \\ f\left( x \right) =\log \sin x \\ f'\left( x \right) =\dfrac { 1 }{ { \sin x } } .\cos x=\cot x \\ \int { { e^{ x } }\left( { \log \sin x+\cot x } \right) dx. } ={ e^{ x } }\left( { \log \sin x } \right) +c \end{array}$$
Hence, this is the answer.
$$\int {\sqrt {1 - \cos x} \,dx = \_\_\_\_\_\_\_ + C;\,2\pi < x < 3\pi } $$
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$$ - 2\sqrt 2 \cos \dfrac{x}{2}$$
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$$ - \sqrt 2 \cos \dfrac{x}{2}$$
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$$ 2\sqrt 2 \cos \dfrac{x}{2}$$
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$$\dfrac{{ - 1}}{2}\sqrt 2 \cos \dfrac{x}{2}$$
Explanation
We have,
$$I= \int { \sqrt { 1-\cos x } dx }$$
$$ I= \int { \sqrt { \left( { { { \sin }^{ 2 } }\dfrac { x }{ 2 } +{ { \cos }^{ 2 } }\dfrac { x }{ 2 } -{ { \cos }^{ 2 } }\dfrac { x }{ 2 } +{ { \sin }^{ 2 } }\dfrac { x }{ 2 } } \right) } dx } $$
$$ I= \int { \sqrt { 2{ { \sin }^{ 2 } }\dfrac { x }{ 2 } } dx } $$
$$I= \sqrt { 2 } \int { \sin \dfrac { x }{ 2 } dx } $$
$$I= \sqrt { 2 } \dfrac{1}{\dfrac{1}{2}}(-\cos \dfrac{x}{2})+C$$
$$I= -2\sqrt { 2 }\cos \dfrac{x}{2}+C$$
Hence, this is the answer.
The integral $$\int \cos \left( \log _ { e } x \right) d x$$ is equal to :
(where $$C$$ is a constant of integration)
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$$\dfrac { { x } }{ 2 } \left[ \sin \left( \log _{ { { e } } }{ x } \right) -\cos \left( \log _{ { { e } } }{ x } \right) \right] +{ C }$$
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$$\dfrac { { x } }{ 2 } \left[ \cos \left( \log _{ { { e } } }{ x } \right) +\sin \left( \log _{ { { e } } }{ x } \right) \right] +{ C }$$
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$${ x }\left[ \cos \left( \log _{ { { e } } }{ x } \right) +\sin \left( \log _{ { { e } } }{ x } \right) \right] +{ C }$$
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$$x \left[ \cos \left( \log _ { e } x \right) - \sin \left( \log _ { c } x \right) \right] + C$$
Explanation
$${ I }=\int \cos (\ell { n }{ x }){ d }{ x }$$
$${ I }=\cos (\ln { } { x })\cdot { x }+\int \sin (\ell { n }{ x }){ d }{ x }$$
$$\cos ( \ell n x ) x + \left[ \sin ( \ell n x ) \cdot x - \int \cos ( \ell n x ) d x \right]$$
$${ I }=\dfrac { { x } }{ 2 } [\sin (\ell { n }{ x })+\cos (\ell { n }{ x })]+{ C }$$
Evaluate: $$\displaystyle\int { \dfrac { { x }^{ 4 }+1 }{ 1+{ x }^{ 6 } } } dx$$
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$${ tan }^{ -1 }(x)-{ tan }^{ -1 }({ x }^{ 3 })+c$$
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$${ tan }^{ -1 }(x)-\frac { 1 }{ 3 } { tan }^{ -1 }({ x }^{ 3 })+c$$
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$${ tan }^{ -1 }(x)+{ tan }^{ -1 }({ x }^{ 3 })+c$$
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$${ tan }^{ -1 }(x)+\frac { 1 }{ 3 } { tan }^{ -1 }({ x }^{ 3 })+c$$
Explanation
Given,
$$\int \dfrac{x^4+1}{x^6+1}dx$$
$$=\int \dfrac{x^4+1}{x^6+1}\times \dfrac{x^2+1}{x^2+1}dx$$
$$=\int \dfrac{(x^6+1)+x^2(x^2+1)}{(x^6+1)(x^2+1)}dx$$
$$=\int \dfrac{dx}{x^2+1}+\dfrac{1}{3}\int \dfrac{3x^2}{x^6+1}dx$$
$$=\tan ^{-1}x+\dfrac{1}{3}3\cdot \int \dfrac{x^2}{x^6+1}dx$$
substitute $$u=x^3$$
$$=\tan ^{-1}x+\dfrac{1}{3}3\cdot \int \dfrac{1}{3\left(u^2+1\right)}du$$
$$=\tan ^{-1}x+\dfrac{1}{3}3\cdot \dfrac{1}{3}\cdot \int \dfrac{1}{u^2+1}du$$
$$=\tan ^{-1}x+\dfrac{1}{3}3\cdot \dfrac{1}{3}\cdot \tan ^{-1}u$$
$$=\tan ^{-1}x+\dfrac{1}{3}3\cdot \dfrac{1}{3}\cdot \tan ^{-1}x^3$$
$$=\tan ^{-1}x+\dfrac{1}{3} \tan ^{-1}x^3+C$$
$$\int { \dfrac { { x }^{ 2 }+1 }{ { x }^{ 4 }+1 } } dx=$$
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$$\dfrac { 1 }{ \sqrt { 2 } } { \tan }^{ -1 }(\dfrac { { x }^{ 2 }+1 }{ \sqrt { 2x } } )+c$$
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$${ \tan }^{ -1 }(\dfrac { { x }^{ 2 }+1 }{ \sqrt { 2x } } )+c$$
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$$\dfrac { 1 }{ \sqrt { 2 } } {\tan }^{ -1 }(\dfrac { { x }^{ 2 }-1 }{ \sqrt { 2x } } )+c$$
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$${ \tan }^{ -1 }(\dfrac { { x }^{ 2 }-1 }{ \sqrt { 2x } } )+c$$
Explanation
Given,
$$\int \dfrac{x^2+1}{x^4+1}dx$$
$$=\int \dfrac{\dfrac{1}{x^2}+1}{\dfrac{1}{x^2}+x^2}dx$$..............dividing by $${x^2}$$
$$=\int \dfrac{\dfrac{1}{x^2}+1}{\dfrac{1}{x^2}+x^2-2+2}dx$$
$$=\int \dfrac{\dfrac{1}{x^2}+1}{\left ( x-\dfrac{1}{x} \right )^2+2}dx$$
$$ x-\dfrac{1}{x}=t\rightarrow dt=\dfrac{1}{x^2}+1 dx$$
$$=\int \dfrac{1}{t^2+2}dt$$
$$=\dfrac{1}{\sqrt{2}}\tan ^{-1}\dfrac{t}{\sqrt 2}+c$$
$$=\dfrac{1}{\sqrt{2}}\tan ^{-1}\dfrac{x-\dfrac{1}{x}}{\sqrt 2}+c$$
$$=\dfrac{1}{\sqrt{2}}\tan ^{-1}\dfrac{x^2-1}{\sqrt 2x}+c$$
$$\int { \frac { dx }{ \sqrt [ 4 ]{ { (x+1) }^{ 5 }{ (x+2) }^{ 3 } } } } $$ is equal to :
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$$4{ (\frac { x+1 }{ x+2 } ) }^{ 1/4 }+c$$
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$$-4{ (\frac { x+1 }{ x+2 } ) }^{ -1/4 }+c$$
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$$-4{ (\frac { x+2 }{ x+1 } ) }^{ 1/4 }+c$$
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None of these
$$\displaystyle\int{{e}^{{\tan}^{-1}{x}}\left(\dfrac{1+x+{x}^{2}}{1+{x}^{2}}\right)dx}$$ is equal to
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$${ -e }^{ { tan }^{ -1 }x }+c$$
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$${ e }^{ { tan }^{ -1 }x }+c$$
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$${ -xe }^{ { tan }^{ -1 }x }+c$$
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$${ xe }^{ { tan }^{ -1 }x }+c$$
Explanation
$$\displaystyle\int{{e}^{{\tan}^{-1}{x}}\left(\dfrac{1+x+{x}^{2}}{1+{x}^{2}}\right)dx}$$
Let $$x=\tan{t}\rightarrow dx={\sec}^{2}{t}dt$$
$$=\displaystyle\int{{e}^{{\tan}^{-1}{\left(\tan{t}\right)}}\left(\dfrac{1+\tan{t}+{\tan}^{2}{t}}{1+{\tan}^{2}{t}}\right){\sec}^{2}{t}dt}$$
$$=\displaystyle\int{{e}^{t}\left(\dfrac{\tan{t}+{\sec}^{2}{t}}{{\sec}^{2}{t}}\right){\sec}^{2}{t}dt}$$
$$=\displaystyle\int{{e}^{t}\left(\tan{t}+{\sec}^{2}{t}\right)dt}$$
$$=\displaystyle\int{{e}^{t}\tan{t}dt}+\displaystyle\int{{e}^{t}{\sec}^{2}{t}dt}$$
$$\tan{t}\displaystyle\int{{e}^{t}dt}-\displaystyle\int{\dfrac{d}{dt}\left(\tan{t}\right)\left(\displaystyle\int{{e}^{t}dt}\right)dt}+\displaystyle\int{{e}^{t}\left({\sec}^{2}{t}\right)dt}+c$$
$$={e}^{t}\tan{t}-\displaystyle\int{{e}^{t}\left({\sec}^{2}{t}\right)dt}+\displaystyle\int{{e}^{t}\left({\sec}^{2}{t}\right)dt}+c$$
$$={e}^{t}\tan{t}+c$$
$$={e}^{{\tan}^{-1}{x}}\tan{{\tan}^{-1}{x}}+c$$
$$=x{e}^{{\tan}^{-1}{x}}+c$$
Hence $$\displaystyle\int{{e}^{{\tan}^{-1}{x}}\left(\dfrac{1+x+{x}^{2}}{1+{x}^{2}}\right)dx}=x{e}^{{\tan}^{-1}{x}}+c$$
$$\int { \cfrac { \cos { x } -1 }{ \sin { x } +1 } } { e }^{ x }dx$$ is equal to:
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$$\cfrac { { e }^{ x }\cos { x } }{ 1+\sin { x } } +c$$
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$$c-\cfrac { { e }^{ x }\sin { x } }{ 1+\sin { x } } $$
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$$c-\cfrac { { e }^{ x }}{ 1+\sin { x } } $$
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$$c-\cfrac { { e }^{ x }\cos { x } }{ 1+\sin { x } } $$
If $$\int { \cfrac { 1 }{ \left( 1+x \right) \sqrt { x } } } dx=f\left( x \right) +A$$, where A is any arbitary constant, then the function f(x) is
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$$2\tan^{-1}{x}$$
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$$2\tan^{-1}{\sqrt{x}}$$
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$$2\cot^{-1}{\sqrt{x}}$$
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$$log_{e}{(1+x)}$$
Explanation
Given,
$$\int \dfrac{1}{\left(1+x\right)\sqrt{x}}dx$$
put $$u=\sqrt{x}$$
$$=\int \dfrac{2}{1+u^2}du$$
$$=2\tan ^{-1}\left(u\right)$$
$$=2\tan ^{-1}\left(\sqrt{x}\right)+C$$
$$=f(x)+A$$
$$\therefore f(x)=2\tan ^{-1}\left(\sqrt{x}\right)$$
$$\int { \dfrac { dx }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+4) } } =$$
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$$\frac { 1 }{ 3 } \tan ^{ -1 }{ x } -\frac { 1 }{ 3 } \tan ^{ -1 }{ \frac { x }{ 2 } } +c$$
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$$\frac { 1 }{ 3 } \tan ^{ -1 }{ x } +\frac { 1 }{ 3 } \tan ^{ -1 }{ \frac { x }{ 2 } } +c$$
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$$\frac { 1 }{ 3 } \tan ^{ -1 }{ x } -\frac { 1 }{ 6 } \tan ^{ -1 }{ \frac { x }{ 2 } } +c$$
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$$\tan ^{ -1 }{ x } -2\tan ^{ -1 }{ \frac { x }{ 2 } } +c$$
Explanation
Given,
$$\int \dfrac{dx}{\left(x^2+1\right)\left(x^2+4\right)}$$
$$=\dfrac 13 \int \dfrac{3 \, dx}{\left(x^2+1\right)\left(x^2+4\right)}$$
$$=\dfrac 13 \int \dfrac{x^2+4-(x^2+1) \, }{\left(x^2+1\right)\left(x^2+4\right)}dx$$
$$=\int \dfrac{1}{3\left(x^2+1\right)}dx-\dfrac{1}{3\left(x^2+4\right)}dx$$
$$=\dfrac{1}{3}\tan ^{-1}\left(x\right)-\dfrac{1}{6}\tan ^{-1}\left(\dfrac{x}{2}\right)+C$$
Integrate: $$\displaystyle \int \dfrac{x}{\sqrt{x+4}}dx$$
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$$\dfrac 23(x+4)^{\tfrac 32}-8\sqrt{x+4}$$
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$$\dfrac 23(x+4)^{\tfrac 32}+8\sqrt{x+4}$$
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$$\dfrac 23(x+4)^{\tfrac 32}+4\sqrt{x+4}$$
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None of these
Explanation
Given,
$$I=\displaystyle \int \dfrac{x}{\sqrt{x+4}}dx$$
let $$u=\sqrt{x+4}$$ $$\Rightarrow du=\dfrac{1}{\sqrt{x+4}}$$
$$\Rightarrow \displaystyle \int \:2\left(u^2-4\right)du$$
$$\displaystyle =2\left(\int \:u^2du-\int \:4du\right)$$
$$=2\left(\dfrac{u^3}{3}-4u\right)$$
$$=2\left(\dfrac{\left(\sqrt{x+4}\right)^3}{3}-4\sqrt{x+4}\right)$$
$$=2\left(\dfrac{1}{3}\left(x+4\right)^{\frac{3}{2}}-4\sqrt{x+4}\right)$$
$$=\dfrac{2}{3}\left(x+4\right)^{\frac{3}{2}}-8\sqrt{x+4}$$
The angle made by the tangent line at (1, 3) on the curve $$y=4x-{ x }^{ 2 }$$ with $$\overline{ OX } $$ is
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$$\tan ^{-1}2$$
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$$\tan ^{-1}(1/2)$$
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$$\tan ^{-1}-2$$
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None of these
Explanation
Given,
$$y=4x-x^2$$
$$\dfrac{dy}{dx}=4-2x$$
$$\dfrac{dy}{dx}_{(1,3)}=4-2(1)=2$$
Therefore, angle made by tangent,
$$\tan \theta =2$$
$$\therefore \theta =\tan ^{-1}2$$
If $$\displaystyle \int { \left( u\cfrac { dv }{ dx } \right) } dx=uv-\int { wdx } $$, then $$w=$$
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$$\cfrac { du }{ dx } \cfrac { dv }{ dx } $$
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$$v\cfrac { du }{ dx } $$
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$$\cfrac { d }{ dx } (uv) $$
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$$u\cfrac { dv }{ dx } $$
Explanation
Given,
$$\displaystyle \int \left ( u\dfrac{dv}{dx} \right )dx=uv-\int wdx$$
it is called as, product rule integration,
$$\displaystyle \int u.v \ dx=u\int v\ dx$$ $$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$$
$$\displaystyle \int \left ( u\dfrac{dv}{dx} \right )dx$$
$$\displaystyle =uv-\int v\dfrac{du}{dx}dx$$
$$\displaystyle \therefore w=v\dfrac{du}{dx}$$
$$\displaystyle \int { \cfrac { x\tan ^{ -1 }{ x } }{ { \left( 1+x^{ 2 } \right) }^{ 3/2 } } } dx=$$
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$$\cfrac { x+\tan ^{ -1 }{ x } }{ { \sqrt { \left( 1+x^{ 2 } \right) } }^{ } } +c$$
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$$\cfrac { x-\tan ^{ -1 }{ x } }{ { \sqrt { \left( 1+x^{ 2 } \right) } }^{ } } +c$$
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$$\cfrac { \tan ^{ -1 }{ x }-x }{ { \sqrt { \left( 1+x^{ 2 } \right) } }^{ } } +c$$
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None of these
Explanation
Let $$I = \displaystyle \int \dfrac{x \tan^{-1} x}{(1 + x^2)^{3/2}} dx $$ let $$x = \tan \theta$$
$$dx = \sec^2 \theta d \theta$$
$$I = \displaystyle \int \dfrac{\tan \theta . \theta . \sec^2 \theta}{\sec^3 \theta} d \theta = \int \dfrac{\theta \tan \theta }{\sec \theta } d \theta$$
$$I = \displaystyle \int \theta. \sin \theta \, d \theta = - \theta \cos \theta + \int \cos \theta $$
$$= - \theta \cos \theta + \sin \theta$$
$$= \dfrac{-\tan^{-1} x}{\sqrt{1 + x^2}} + \dfrac{x}{\sqrt{1 + x^2}} = \dfrac{x - \tan^{-1} x}{\sqrt{1 + x^2}} + C$$
$$\displaystyle \int \dfrac{dx}{9x^2 + 1} =$$_____.
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$$\dfrac{1}{3} \tan^{-1} (2x) + c$$
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$$\dfrac{1}{3} \tan^{-1} x + c$$
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$$\dfrac{1}{3} \tan^{-1} (3x) + c$$
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$$\dfrac{1}{3} \tan^{-1} (6x) + c$$
Explanation
Consider the given integral.
$$I=\displaystyle \int \dfrac{1}{(9x^2+1)}dx$$
$$I=\displaystyle \int \dfrac{1}{9(x^2+\dfrac{1}{9})}dx$$
$$I=\dfrac{1}{9}\displaystyle \int \dfrac{1}{x^2+(\dfrac{1}{3})^2}dx$$
$$I=\dfrac{1}{9}\times \dfrac{1}{1/3}\tan^{-1}(\dfrac{x}{1/3})+C$$
$$I=\dfrac{1}{3} \tan ^{-1}{3x}+C$$
Hence, this is the answer.
Evaluate $$\displaystyle \int \:e^x\left(\log \left(x\right)+\dfrac{1}{x^2}\right)dx$$
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$$e^x(\log{x}+\cfrac{1}{x^2})$$
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$$e^x(\log{x}+\cfrac{1}{x})$$
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$$e^x(\log{x}-\cfrac{1}{x^2})$$
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$$e^x(\log{x}-\cfrac{1}{x})$$
Explanation
Given,
$$\displaystyle \int \:e^x\left(\log \left(x\right)+\dfrac{1}{x^2}\right)dx$$
$$\displaystyle =\int \:e^x\log \left(x\right)dx+\int \dfrac{e^x}{x^2}dx$$
We know,
$$\displaystyle \int u.v \ dx=u\int v\ dx$$ $$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$$
$$\displaystyle =\left (e^x\log \left(x\right)-\text{Ei}\left(x\right) \right )-\dfrac{e^x}{x}+\text{Ei}\left(x\right)$$ where $$\displaystyle\text{Ei}(x)=\int \dfrac{e^x}{x} dx$$
$$\displaystyle =e^x\log \left(x\right)-\dfrac{e^x}{x}+C$$
$$=e^x(\log x-\dfrac{1}{x})$$
$$\int { \left( 2+\log { x } \right) { \left( ex \right) }^{ x }dx } =.....+C;x>1$$
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$${ \left( ex \right) }^{ x }$$
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$${ x }^{ x }$$
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$${ \left( ex \right) }^{ -x }$$
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$${ e }^{ { x }^{ x } }$$
Explanation
Let $${ \left( ex \right) }^{ x }=t$$
$$x\ln { ex } =\ln { t } $$
$$\Rightarrow x(1+\ln { x } )=\ln { t } $$
$$\Rightarrow \left( x.\cfrac { 1 }{ x } +\left( 1+\ln { x } \right) \right) dx=\cfrac { 1 }{ t } dt$$
$$\Rightarrow { \left( ex \right) }^{ x }\left( 2+\ln { x } \right) dx=dt$$
now
integrate on both sides
$$\int { dt } =t+c\Rightarrow { \left( ex \right) }^{ x }+c$$
If $$\displaystyle \int { e^ { \sec} } (\sec x\, \tan x \, f(x)+ \sec x\, \tan x+ \tan^2x)dx=e^{\sec x}f(x)+c. Then f(x) is$$
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$$\sec x+x\ tan x+\dfrac{1}{2}$$
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$$x \sec x+x \tan x+\dfrac{1}{2}$$
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$$x \sec x+x^2 \tan x+\dfrac{1}{2}$$
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$$\sec x+ \tan x+\dfrac{1}{2}$$
Explanation
$$\displaystyle \int { e^ { secx} } (\sec x\, \tan x \, f(x)+ \sec x \, \tan x+ \sec^2x)dx=e^{sec x}f(x)$$
Differentiate both side
$$\int { e^ { secx} } (\sec x\, \tan x\, f(x)+\sec x\,\tan x+\sec^2x)dx=e^{secx}f'(x)+e^{secx}\,tan x\,secx\,f(x)$$
$$\therefore \sec x\,tanx+\sec^2x=f'(x)$$
$$\therefore f(x)=\sec x+\tan x+d$$
$$\int _{ a }^{ b }{ \dfrac{dx}{2 + 3x} } =$$
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$$\dfrac{1}{3} log_e (2 + 3b)$$
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$$\dfrac{1}{3} log_e (2 + 2a)$$
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$$\dfrac{1}{3} log_e \left(\dfrac{2 + 3b}{2 + 3a}\right)$$
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$$\dfrac{1}{3} log_e \left(\dfrac{2 + 3a}{2 + 3b}\right)$$
$$\displaystyle\int \dfrac{2x^3-1}{x^4+x}dx$$ is equal to?
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$$ln\left|\dfrac{x^3+1}{x}\right|+c$$
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$$ln\left|\dfrac{x^3+1}{x^2}\right|+c$$
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$$\dfrac{1}{2}\ln\left|\dfrac{x^3+1}{x^2}\right|+c$$
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$$\dfrac{1}{2}ln\left|\dfrac{x^3+1}{x}\right|+c$$
Explanation
$$\displaystyle\int \dfrac{2x^3-1}{x^4+x}dx=\displaystyle\int \dfrac{2x-x^{-2}}{x^2+x^{-1}}dx=ln(x^2+x^{-1})+c=ln(x^3+1)-lnx+c$$.
If $$\displaystyle\int 2^{2x}\cdot 2^xdx=A\cdot 2^{2^x}+c$$, then $$A=?$$
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$$\dfrac{1}{log 2}$$
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$$log 2$$
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$$(log 2)^2$$
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$$\dfrac{1}{(log 2)^2}$$
Explanation
$$2^x=z\Rightarrow 2^xdx=\dfrac{dz}{ln 2}$$
$$\Rightarrow \dfrac{1}{ln 2}\displaystyle\int 2^zdz=\dfrac{1}{(ln 2)^2}2^{2^x}+c$$.
If $$\displaystyle \int \dfrac{dx}{x^3(1+x^6)^{2/3}} = xf(x) (1 + x^6)^{\frac{1}{3}} + C$$
where $$C$$ is a constant of integration, then the function $$f(x)$$ is equal to -
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$$-\dfrac{1}{6x^3}$$
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$$\dfrac{3}{x^2}$$
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$$-\dfrac{1}{2x^2}$$
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$$-\dfrac{1}{2x^3}$$
Explanation
$$\displaystyle \int \dfrac{dx}{x^3(1+x^6)^{2/3}} = xf(x) ( 1 +x^6)^{1/3} + C$$
$$\displaystyle \int \dfrac{dx}{x^7\left(\dfrac{1}{x^6} + 1\right)^{2/3}} = xf(x) (1 + x^6)^{1/3} + C$$
Let $$t = \dfrac{1}{x^6} + 1$$
$$dt = \dfrac{-6}{x^7} dx$$
$$\displaystyle -\dfrac{1}{6} \int \dfrac{dt}{t^{2/3}} = - \dfrac{1}{2} t^{1/3}$$
$$=-\dfrac{1}{2} \left(\dfrac{1
}{x^6} + 1\right)^{1/3} = -\dfrac{1}{2} \dfrac{(1+x^6)^{1/3}}{x^2}$$
$$\therefore f(x) = -\dfrac{1}{2x^3}$$
Evaluate : $$ \int \dfrac { sec x}{( sec x + tan x ) } dx $$
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$$ tan x+ sec x+ C $$
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$$ tan x -sec x + C $$
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$$ -tan x + sec x + C $$
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$$ -tan x - sec x + C $$
What is $$\displaystyle \int \dfrac{dx}{2x^2 - 2x + 1}$$ equal to ?
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$$\dfrac{\tan^{-1} (2x - 1)}{2} + c$$
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$$2 \tan^{-1} (2x - 1) += c$$
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$$\dfrac{\tan^{-1} (2x + 1)}{2} + c$$
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$$\tan^{-1} (2x - 1) + c$$
Explanation
Given,
$$\int \dfrac{1}{2x^2-2x+1}dx$$
complete the square
$$=\int \dfrac{1}{2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}}dx$$
apply $$u=x-\frac{1}{2}$$
$$\int \dfrac{2}{4u^2+1}du$$
apply $$u=\frac{1}{2}v$$
$$=2\cdot \int \dfrac{1}{2\left(v^2+1\right)}dv$$
$$=2\cdot \dfrac{1}{2}\tan ^{-1} \left(v\right)$$
$$=\tan ^{-1} \left(2\left(x-\dfrac{1}{2}\right)\right)$$
$$=\tan ^{-1} \left(2x-1\right)$$
$$=\tan ^{-1}\left(2x-1\right)+C$$
$$\displaystyle \int\frac{x\cos x\log x-\sin x}{x(\log x)^{2}}dx=$$
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$$\displaystyle \frac { \sin { x } }{ \log { x } } +C$$
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$$\displaystyle \frac { \cos { x } }{ \log { x } } +C$$
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$$\displaystyle \frac { \log { x } }{ \sin { x } } +C$$
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$$\displaystyle \frac { \log { x } }{ \cos { x } } +C$$
Explanation
$$\displaystyle\int { \frac { x\cos(x)\log(x)-\sin(x) }{ x{ (\log(x)) }^{ 2 } } dx }$$
$$\displaystyle=\int { \frac {\cos(x) }{\log(x) } -\frac {\sin(x) }{ x{ (\log(x)) }^{ 2 } } } dx$$
$$\displaystyle=\int { \frac { \cos(x) }{ \log(x) } dx } -\int { \frac { \sin(x) }{ x{ (\log(x)) }^{ 2 } } dx }$$
$$\displaystyle=\frac { 1 }{ \log(x) } \sin(x)-\int { \frac { 1 }{ x{ (\log(x)) }^{ 2 } } (-\cos(x)) } dx-\int { \frac { \sin(x) }{ x{ (\log(x)) }^{ 2 } } dx } +C$$
$$\displaystyle=\frac { \sin(x) }{ \log(x) } +\int { \frac {\ sin(x) }{ x{ (\log(x)) }^{ 2 } } dx } -\int { \frac { \sin(x) }{ x{ (\log(x)) }^{ 2 } } dx } +C$$
$$\displaystyle=\frac { \sin(x) }{ \log(x) } +C\\ $$
If $$\int { x\sin { x } } dx=-x\cos { x } +\alpha $$, then $$\alpha$$ is equal to
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$$\sin x+C$$
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$$\cos x+C$$
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$$C$$
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None of these
Explanation
$$I=\displaystyle\int{x\sin{x}dx}$$
Integrating by parts,
Let $$u=x\Rightarrow\,du=dx$$
$$dv=\sin{x}dx\Rightarrow\,v=-\cos{x}$$
$$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ] $$......by parts formula
$$I=-x\cos{x}+\displaystyle\int{\cos{x}dx}$$
$$I=-x\cos{x}+\sin{x}+c$$ ......where $$c$$ is the constant of integration.
We have $$I=\displaystyle\int{x\sin{x}dx}=-x\cos{x}+\alpha$$
Comparing with $$I=-x\cos{x}+\sin{x}+c$$ we have
$$\alpha=\sin{x}+c$$
Evaluate : $$\displaystyle \int \dfrac { cos 2x}{cos^2 x sin^2 x }dx $$
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$$ - cot x -tan x +C $$
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$$ -cot x + tan x + C$$
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$$ cot x -tan x +C $$
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$$ cot x + tan x + C$$
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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