The value of $\int \frac{(x-1) \mathrm{e}^x}{(x+1)^3} \mathrm{~d} x$ is equal to

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } \mathrm{I}=\int \frac{(x-1) \mathrm{e}^x}{(x+1)^3} \mathrm{~d} x \ & \begin{array}{l}=\int \frac{(x+1-2) \mathrm{e}^x}{(x+1)^3} \mathrm{~d} x \ =\int \mathrm{e}^x\left[\frac{1}{(x+1)^2}-\frac{2}{(x+1)^3}ight] \mathrm{d} x \ =\frac{\mathrm{e}^x}{(x+1)^2}+\mathrm{c} \ \quad \ldots\left[\because \int \mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)ight] \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}ight]\end{array}\end{aligned}$

The value of $\int \frac{(x-1) \mathrm{e}^x}{(x+1)^3} \mathrm{~d} x$ is equal to

Solution