Mathematics

$\cfrac{1}{1+x}e^x+c$

##### SOLUTION
Let $I=\int{\cfrac{xe^x}{(1+x)^2}dx}$
$I=\int{\cfrac{xe^x+e^x-e^x}{(1+x)^2}dx}$
$I=\int{\cfrac{(x+1)e^x}{(1+x)^2}dx}$ $+\int{\cfrac{e^x}{(1+x)^2}dx}$
$I=\int{\cfrac{e^x}{(1+x)}dx}$ $+\int{\cfrac{e^x}{(1+x)^2}dx}$
Let $f(x)=\cfrac{1}{1+x}$
So, $g(x)=f'(x)=\cfrac{-1}{(1+x)^2}$
Using identity
$I=\int{f(x)e^xdx}+$$\int{f'(x)e^xdx}=f(x)e^x$
So, $I=\cfrac{1}{1+x}e^x+c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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