Evaluate the following integrals: ^ x (1-x)^2 (1+x^2)^2 dx - Vidyadip

Question

Evaluate the following integrals:

$\int\text{e}^{\text{x}}\frac{(1-\text{x})^2}{(1+\text{x}^2)^2}\text{dx}$

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Answer

Let $\text{I}=\int\text{e}^{\text{x}}\bigg[\frac{(1-\text{x})^2}{(1+\text{x}^2)^2}\bigg]\text{dx}$
$=\int\text{e}^\text{x}\bigg[\frac{1+\text{x}^2-2\text{x}}{(1+\text{x}^2)^2}\bigg]\text{dx}$
$=\int\text{e}^{\text{x}}\bigg[\frac{1}{1+\text{x}^2}-\frac{2\text{x}}{(1+\text{x}^2)^2}\bigg]\text{dx}$
Here, $\text{f(x)}=\frac{1}{1+\text{x}^2}$
$\Rightarrow\text{f}'(\text{x})=\frac{-2\text{x}}{(1+\text{x}^2)^2}$
Put $\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\frac{1}{1+\text{x}^2}=\text{t}$
Diff. both sides w.r.t w
$\text{e}^{\text{x}}\frac{1}{1+\text{x}^2}+\text{e}^{\text{x}}\frac{-1}{(1+\text{x}^2)^2}2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}\bigg[\frac{1}{1+\text{x}^2}-\frac{2\text{x}}{(1+\text{x}^2)^2}\bigg]\text{dx = dt}$
$\therefore\int\text{e}^{\text{x}}\bigg[\frac{1}{1+\text{x}^2}-\frac{2\text{x}}{(1+\text{x}^2)^2}\bigg]\text{dx}=\int\text{dt}$
$\Rightarrow\text{I}=\text{t}+\text{C}$
$=\frac{\text{e}^{\text{x}}}{1+\text{x}^2}+\text{C}$

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