Question

Evaluate:

$\int e^x[{\tan }^{-1}x+\frac{1}{1+x^2}]dx$.

Answer 1

Now,

$\int e^x[{\tan }^{-1}x+\frac{1}{1+x^2}]dx$

$=\int e^x.{\tan }^{-1}xdx$$+\int \frac{e^x}{1+x^2}dx$

$=e^x.{\tan }^{-1}xdx-\int \frac{e^x}{1+x^2}dx$$+\int \frac{e^x}{1+x^2}dx$ [ Using method of by parts]

$=e^x{\tan }^{-1}x+c$. [ Where $c$ is integrating constant]