Mathematics

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

##### SOLUTION

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

Put $x + 1= u$ and $x^2= (u-1)^2$

$\implies dx = du$
Therefore, $C = e^{-1} \int \dfrac{( (u-1)^2 + 1) e^u}{u^2} du$

$= e^{-1} \int \dfrac{( u^2+1-2u + 1) e^u}{u^2} du$

$I = e^{-1} \int e^u du + e^{-1} \int \dfrac{ (2 – 2u ) e^u}{u^2} du$

$I = e^{-1} e^u + 2 e^{-1} \int ( \dfrac{1}{u^2} e^u - \dfrac{1}{u} e^u) du$

$I = e^{u-1} + 2e^{-1} \int \dfrac{1}{u^2} e^u du – 2^{-1} \dfrac{1}{u} e^u + 2e^{-1} \int - \dfrac{1}{u^2} e^u du$

$I = e^{u-1} - \dfrac{2 e^{u-1}}{u} + c$

$I = e^{x} -\dfrac{ 2e^x}{x+1} + c$

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

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

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

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