Question

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

Answer 1

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