Question

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

Answer 1

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