Mathematics

# Evaluate $\int \dfrac {(x - 1)e^{x}}{(x + 1)^{3}} dx$.

$\dfrac {e^{x}}{(x + 1)^{2}} + C$

##### SOLUTION
Let $I =\displaystyle \int \dfrac {(x - 1)e^{x}}{(x + 1)^{3}} dx$
$\Rightarrow I =\displaystyle \int \left \{\dfrac {x + 1 - 2}{(x + 1)^{3}}\right \} e^{x} dx$
$=\displaystyle \int \left \{\dfrac {1}{(x + 1)^{2}} - \dfrac {2}{(x + 1)^{3}}\right \}e^{x} dx$
$= \displaystyle \int e^{x}\cdot \dfrac {1}{(x + 1)^{2}} dx - 2\int e^{x} \dfrac {1}{(x + 1)^{3}} dx$
Applying integrating by parts, we get
$=\displaystyle \left (\dfrac {1}{(x + 1)^{2}} e^{x} - \int e^{x} \dfrac {(-2)}{(x + 1)^{3}} dx\right ) - 2\int e^{x} \dfrac {1}{(x + 1)^{3}}dx$
$= \dfrac {e^{x}}{(x + 1)^{2}} + C$
Note We can use the formula
$\int e^{x} [f(x) + f'(x)] dx = e^{x}f (x) + C$.

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

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