Question

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

Answer 1

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