Question

$\int e^x\frac{x-1}{{\left(x+1\right)}^3}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int e^x\left[\frac{x-1}{{\left(x-1\right)}^3}\right]dx \\ =\int e^x\left[\frac{x+1-2}{{\left(x+1\right)}^3}\right]dx \\ =\int e^x\left[\frac{1}{{\left(x-1\right)}^2}-\frac{2}{{\left(x+1\right)}^3}\right]dx \\ \mathrm{Here},f\left(x\right)=\frac{1}{{\left(x+1\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)=\frac{-2}{{\left(x+1\right)}^2} \\ \mathrm{Put}{e}^{x}f\left(x\right)=t \\ \mathrm{let}{e}^x\frac{1}{{\left(x+1\right)}^2}=t \\ \mathrm{Diff}\mathrm{both}\mathrm{sides} \\ {e}^x\frac{1}{{\left(x+1\right)}^2}+e^x\frac{\left(-2\right)}{{\left(x+1\right)}^3}=\frac{dt}{dx} \\ \Rightarrow e^x\left[\frac{1}{{\left(x+1\right)}^2}-\frac{2}{{\left(x+1\right)}^3}\right]dx=dt \\ \therefore \int e^x\left[\frac{1}{{\left(x+1\right)}^2}-\frac{2}{{\left(x+1\right)}^3}\right]dx=\int dt \\ =t+C \\ =\frac{e^x}{{\left(x+1\right)}^2}+C \end{gathered}$$