Mathematics

# Evaluate $\displaystyle \int \frac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}} dx$

##### SOLUTION
Integrating by parts we get,
$u = xe^x \space ,v = \cfrac{1}{(1+x)^2}$
$u^{'} = e^x+xe^x , \int v dx = \cfrac{-1}{1+x}$
$\int \cfrac{xe^x}{(1+x)^2} dx = -\cfrac{xe^x}{1+x} - \int \cfrac{-e^x(x+1)}{1+x}dx$
$\int \cfrac{xe^x}{(1+x)^2} dx= -\cfrac{xe^x}{1+x} + \int e^x dx$
$\int \cfrac{xe^x}{(1+x)^2} dx= -\cfrac{xe^x}{1+x} + e^x$
$\int \cfrac{xe^x}{(1+x)^2} dx= \cfrac{e^x}{1+x} + C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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