Mathematics

# The value of $\int { { e }^{ x } } .\dfrac { { x }^{ 2 }+1 }{ { \left( x+1 \right) }^{ 2 } } dx$ is

##### ANSWER

${ e }^{ x }\left( \dfrac { x-1 }{ x+1 } \right) +C$

##### SOLUTION
$I=\displaystyle\int e^x\dfrac{x^2+1}{(x+1)^2}dx$
$=\displaystyle\int e^x\dfrac{x^2+1+2x-2x}{(x+1)^2}dx$
$=\displaystyle\int e^x\dfrac{(x+1)^2-2x}{(x+1)^2}dx$
$=\displaystyle\int e^x\left\{1-\dfrac{2x}{(x+1)^2}\right\}dx$
$=\displaystyle\int e^xdx-\displaystyle\int e^x\cdot \dfrac{2x}{(x+1)^2}dx$.
$=e^x-I_1+e$
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 $\displaystyle\int e^x[f(x)+f'(x)]dx$
$=2e^x\cdot \left(\dfrac{1}{x+1}\right)$   $[\because\displaystyle\int e^x[f(x)+f'(x)]dx=e^xf(x)+c]$
$=\dfrac{2e^x}{x+1}$
$\therefore I=e^x-\dfrac{2e^x}{x+1}+c$
$=e^x\left(1-\dfrac{2}{x+1}\right)+c$
$=e^x\left\{\dfrac{x+1-2}{x+1}\right\}+c$
$\therefore I=e^x\left(\dfrac{x-1}{x+1}\right)+c$.

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

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