Evaluate : (x^2+1 ) e^x (x+1)^2 d x - Vidyadip

Question

Evaluate : $\int \frac{\left(x^2+1\right) \cdot e^x}{(x+1)^2} \cdot d x$

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Answer

$
\begin{aligned}
\text { I } & =\int e^x\left[\frac{x^2+1}{(x+1)^2}\right] \cdot d x \\
& =\int e^x\left[\frac{x^2-1+2}{(x+1)^2}\right] \cdot d x \\
& =\int e^x\left[\frac{x^2-1}{(x+1)^2}+\frac{2}{(x+1)^2}\right] \cdot d x \\
& =\int e^x\left[\frac{x-1}{x+1}+\frac{2}{(x+1)^2}\right] \cdot d x
\end{aligned}
$
Here $f(x)=\frac{x-1}{x+1}$
$
\begin{aligned}
\Rightarrow & f^{\prime}(x)=\frac{(x+1)(1)-(x-1)(1)}{(x+1)^2}=\frac{2}{(x+1)^2} \\
& \because \quad \int\left[f(x)+f^{\prime}(x)\right] \cdot d x=e^x \cdot f(x)+c \\
\text { I } & =e^x \cdot\left(\frac{x-1}{x+1}\right)+c \\
\therefore & \int \frac{\left(x^2+1\right) \cdot e^x}{(x+1)^2} \cdot d x=e^x \cdot\left(\frac{x-1}{x+1}\right)+c
\end{aligned}
$

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