Finding value of $\int\frac{\ln(x)}{1+x^2}dx$ 
Finding value of $\displaystyle \int\frac{\ln(x)}{1+x^2}dx$

Try: let $$I=\int\frac{\ln(x)}{1+x^2}dx=\frac{1}{2}\int\bigg[\frac{\ln x}{1-ix}+\frac{\ln x}{1+ix}\bigg]dx$$
$$I=\frac{1}{2}\int\frac{\ln x}{1-ix}dx+\frac{1}{2}\int\frac{\ln x}{1+ix}dx$$
Put $1+ix =t$ Then $dx=-idt$ and $1-ix=u$ Then $dx=idu$
So $$I=\frac{i}{2}\int \frac{\ln(1-u)-\ln(i)}{u}du-\frac{-i}{2}\int \frac{\ln(1-t)-\ln(i)}{t}dt$$
Could some help me to solve it , Thanks

Note that
$$\ln(x)+C=\int\frac1x{\rm~d}x$$
and
$${\rm Li}_2(x)+C=\int\frac{\ln(1-x)}x{\rm~d}x$$
where ${\rm Li}$ is the polylogarithm, since
$${\rm Li}_1(x)=\ln(1-x)$$
and
$${\rm Li}_{s+1}(x)=\int_0^x\frac{{\rm Li}_s(t)}t{\rm~d}t$$
