$\frac{\left(\tan ^{-1} x\right)^2}{4}-\frac{x \tan ^{-1} x}{2\left(1+x^2\right)}+\frac{1-x^2}{4\left(1+x^2\right)}+C$

$\frac{\left(\tan ^{-1} x\right)^2}{4}-\frac{4 x \tan ^{-1} x+1-x^2}{8\left(1+x^2\right)}+C$

$\frac{\left(\tan ^{-1} x\right)^2}{4}-\frac{x \tan ^{-1} x}{\left(1+x^2\right)}-\frac{1-x^2}{4\left(1+x^2\right)}+C$

$\frac{(\tan x)^2}{4}+\frac{4 x \tan ^{-1} x-1+x^2}{4\left(1+x^2\right)}+C$

$\frac{1}{2}\left[\frac{1}{1+x}+\frac{\log x}{(1+x)^2}-\log \left(x^2+x\right)\right]+C$

$\frac{1}{2}\left[\frac{1}{1+x}-\frac{\log x}{(1+x)}-\log \left(1+x^2\right)\right]+C$

$\frac{1}{2}\left[\frac{1}{1+x}+\frac{\log x}{(1+x)^2}-\log \left(1+x^2\right)\right]+C$

$\frac{1}{2}\left[\frac{1}{1+x}-\frac{\log x}{(1+x)^2}+\log \left(\frac{x}{1+x}\right)\right]+C$

$\frac{8}{5}$

$\frac{16}{5}$

$\frac{3}{5}$

$\frac{19}{5}$

$\frac{(\sin x-\cos x) x-\sin x \cos x}{x \sin x \cos x}+C$

$-\frac{1}{x}+\frac{\sin x+\cos x}{\cos x-\sin x}+c$

$-\frac{1}{x}+\frac{\sin x-\cos x}{\sin ^2 x \cos ^2 x}+C$

$\frac{(\sin x-\cos x) x-\sin x-\cos x}{x(\sin x+\cos x)}+C$