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

Question

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

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Answer

Let
$
\begin{aligned}
I & =\int e^x\left(\frac{1-x}{1+x}\right)^2 d x \\
I & =\int \frac{e^x(1-x)^2}{(1+x)^2} d x=\int e^x\left(\frac{(1+x)^2-4 x}{(1+x)^2}\right) d x \\
& =\int e^x d x-4 \int \frac{x e^x}{(1+x)^2} d x \\
& =\int e^x d x-4 \int e^x \frac{(x+1-1)}{(1+x)^2} d x \\
& =\int e^x d x-4 \int\left(\frac{1}{(1+x)}-\frac{1}{(1+x)^2}\right) e^x d x \\
& =e^x-4 \int\left(\frac{1}{1+x}-\frac{1}{(1+x)^2}\right) e^x d x
\end{aligned}
$
Let $\quad f(x)=\frac{1}{1+x} \therefore f^{\prime}(x)=\frac{-1}{(1+x)^2}$
So it is in the form of $\int\left(f(x)+f^{\prime}(x)\right) e^x d x$.
So its value is equal to $e^x f(x)+C$.
$
\begin{aligned}
I & =e^x-4\left(e^x \times \frac{1}{1+x}\right)+C \\
& \quad\left(\because f(x)=\frac{1}{1+x}\right) \\
& =e^x\left(\frac{1+x-4}{1+x}\right)+C=\left(\frac{x-3}{x+1}\right) e^x+C \text { Ans. }
\end{aligned}
$

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