evaluate $\int \frac{\tan x}{x^2+1}\:dx$

Question

$$\int \frac{\tan x}{x^2+1}\, \mathrm dx$$

I used By-parts method setting $u=\tan x$ and $\, \mathrm dv=\frac{1}{x^2+1}\, \mathrm dx$, but then I got an integral that's more complicated

I also thought of trigonometric substitution, setting $x=\tan\theta$, but how am going to substitute that for the $\tan x$ in numerator?

I tried to use websites like symbolab & wolfram to evaluate the integral but I got no result.

Answer 1

Hope it helps. It also turns into a more complicated as I thought. You might use Matlab to calculate this.

Answer 2

The Laurent series of tan(x) is $$\sum_{n=1,3,5..}^{\infty }\frac{8x}{(n\pi )^2-4x^2}$$ so $$\frac{\tan(x)}{1+x^2}=\sum_{n=1,3,5..}^{\infty }\frac{8x}{\left [(n\pi )^2-4x^2 \right ](1+x^2)}$$

use the partial fraction to get $$\sum_{n=1,3,5,..}^{\infty }\frac{8x}{((n\pi)^2+4 )(1+x^2)}+\frac{8}{((n\pi)^2+4 )(n\pi -2x)}-\frac{8}{((n\pi)^2+4 )(n\pi +2x)}$$

$$\int \frac{\tan x}{1+x^2}dx=C+\sum_{n=1,3,5,..}^{\infty }\frac{4}{(n\pi )^2+4}\left [ \log(1+x^2)-\log(n\pi -2x)-\log(n\pi +2x) \right ]$$

hence

$$\int \frac{\tan x}{1+x^2}dx=C+\sum_{n=1,3,5,..}^{\infty }\frac{4}{(n\pi )^2+4}\left [ \log(\frac{1+x^2}{(n\pi )^2-4x^2}) \right ]$$