$\int\left(\frac{x+2}{x+4}\right)^2 \cdot e^x d x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{I}=\int\left(\frac{x+2}{x+4}ight)^2 \mathrm{e}^x \mathrm{~d} x \ &=\int\left(\frac{x^2+4 x+4}{(x+4)^2}ight) \mathrm{e}^x \mathrm{~d} x \ &=\int \mathrm{e}^x\left[\frac{x(x+4)}{(x+4)^2}+\frac{4}{(x+4)^2}ight] \mathrm{d} x \ &=\int \mathrm{e}^x\left[\frac{x}{(x+4)}+\frac{4}{(x+4)^2}ight] \mathrm{d} x \ &=\mathrm{e}^x\left(\frac{x}{x+4}ight)+\mathrm{c} \ & \quad \cdots\left[\int\left(\mathrm{f}(x)+\mathrm{f}^{\prime}(x) \mathrm{e}^x \mathrm{~d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}ight)ight]\end{aligned}$

$\int\left(\frac{x+2}{x+4}\right)^2 \cdot e^x d x=$

Solution