Mathematics

# Simplify: $\int {\dfrac{{{e^x}\left( {x - 1} \right)}}{{{{\left( {x + 1} \right)}^3}}}dx}$

##### SOLUTION
$I=\int e^x\frac{(x-1)}{(x+1)^3}dx$
$I=\int e^x\frac{(x+1-2)}{(x+1)^3}dx$
$I=\int e^x\frac{1+x}{(x+1)^3}dx$ $-$ $\int e^x\frac{2}{(1+x)^3}dx$
$I=\int e^x\frac{1}{(1+x)^2}dx$ $-$ $\int e^x\frac{2}{(x+1)^3}dx$
By integration by parts : $\int [f(x) g(x)]dx=f(x)\cdot \int g(x)dx-\int[f'(x) \int g(x) dx]dx$
By integration by  parts of first integral
$I=$ $\frac { 1 }{ (1+ x )^{ 2 } } \int { e } ^{ x }-\int { \{ (\frac { d(\frac { 1 }{ (1+ x )^{ 2 } } ) }{ dx } } )\int { { e }^{ x } } dx\} dx$   $-$ $\int e^x\frac{2x}{(1+x^2)^2}dx$
$I= e^x\frac{1}{(1+x)^2}$ $+$ $\int e^x\frac{2}{(1+x)^3}dx$ $-$ $\int e^x\frac{2}{(1+x)^3}dx$
$I= e^x\frac{1}{(1+x)^2}$ $+$ $C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Subjective Medium

Evaluate the following definite integral:

$\displaystyle\int_{0}^{2}x\sqrt{x+2}dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Solve : $\int \dfrac{cos 2x \, sin 4x}{cos^4 x (1 + cos^2 2x)} dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\frac { d } { d x } f ( x ) = g ( x )$ then the value of $\int _ { a } ^ { b } f ( x ) g ( x ) d x$ is
• A. $f ( b ) - f ( a )$
• B. $g ( b ) - g ( a )$
• C. $\frac { 1 } { 2 } \left[ \{ f ( b ) \} ^ { 2 } - \{ f ( a ) \} ^ { 2 } \right]$
• D. $\frac { 1 } { 2 } \left[ \{ g ( b ) \} ^ { 2 } - \{ g ( a ) \} ^ { 2 } \right]$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
Evaluate: $\displaystyle \int \dfrac{\left ( \sec x\:co\sec \right )}{\left ( \log \tan x \right )} dx$
• A. $\displaystyle \log \left ( \tan x \right )$
• B. $\displaystyle \cot \left ( \log x \right )$
• C. $\displaystyle \tan \left ( \log x \right )$
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1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$