Mathematics

# $\displaystyle \int {{e^x}\dfrac{{x - 1}}{{{{\left( {x + 1} \right)}^3}}}{\text{dx}}\;{\text{equals}}}$

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

##### SOLUTION
Solution:-

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

$= \int e^{x}. \dfrac{x+1-2}{(x+1)^{3}}dx$

$= \int e^{x}.\left \{ \dfrac{x+1}{(x+1)^{3}}-\dfrac{2}{(x+1)^{3}} \right \}dx$
$= \int \left \{ \dfrac{e^{x}}{(x+1)^{2}}-\dfrac{2e^{x}}{(x+1)^{3}} \right \}dx$

Let $\dfrac{e^{x}}{(x+1)^{2}} = z$

$\therefore \left \{ \dfrac{e^{x}(x+1)^{2}-x^{2}.2(2x+1)}{(x+1)^{4}} \right \}dx = dz$

$\left \{ \dfrac{e^{x}}{(x+1)^{2}}-\dfrac{2e^{x}}{(x+1)^{3}} \right \}dx = dz$

$\therefore I = \int dz = z+c$

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

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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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