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
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