Integrate $\int_{0}^{\infty}\frac{\tan^{-1}x}{x\left(1+x^{2}\right)}dx$ Q:

Integrate $I=\int_{0}^{\infty}\frac{\tan^{-1}x}{x\left(1+x^{2}\right)}dx$

My Approach:
Put $$\tan^{-1}x=t\to x=\tan t$$
Also we have, $$\frac{dx}{1+x^{2}}=dt$$
We get, $$I=\int_{0}^{\frac{\pi}{2}}\frac{t}{\tan t}dt$$
I'm stuck here, how do I proceed further? I tried integration by parts but it doesn't seem to work out for me.
Edit: I tried again and I got the answer 

Another way using Feynmann trick
$$\frac d {da}\int_0^\infty \frac{\tan ^{-1}(a x)}{x \left(x^2+1\right)}\,dx=\int_0^\infty \frac{dx}{\left(x^2+1\right) \left(a^2 x^2+1\right)}$$
$$\frac{1}{\left(x^2+1\right) \left(a^2 x^2+1\right)}=\frac{a^2}{\left(a^2-1\right) \left(a^2x^2+1\right)}-\frac{1}{\left(a^2-1\right) \left(x^2+1\right)}$$
$$\int \frac{dx}{\left(x^2+1\right) \left(a^2 x^2+1\right)}=\frac{a \tan ^{-1}(a x)-\tan ^{-1}(x)}{a^2-1}$$
Assuming $a>0$
$$\int_0^\infty \frac{dx}{\left(x^2+1\right) \left(a^2 x^2+1\right)}=\frac \pi{2(1+a)}$$ Integrating from $0$ to $1$ then
$$\int_0^\infty \frac{\tan ^{-1}( x)}{x \left(x^2+1\right)}\,dx=\frac{1}{2} \pi  \log (2)$$ and making it more general
$$\int_0^\infty \frac{\tan ^{-1}(k x)}{x \left(x^2+1\right)}\,dx=\frac{1}{2} \pi  \log (k+1)$$
