Question

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

Answer 1

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