How to calculate $\int_0^{\infty}\frac{\ln(x)}{1+x^2}\ \mathrm dx$ 
How does one go about calculating :
$$\int_0^{\infty}\frac{\ln x}{1+x^2}dx$$

I've tried Integration by parts, and failed over and over again
 A: $$\int_0^{\infty}\frac{\ln(x)}{1+x^2}dx=\color{#C00000}{\int_0^{1}\frac{\ln(x)}{1+x^2}dx}+\int_1^{\infty} \frac{\ln(x)}{1+x^2}dx$$

Let's find out :
$$\color{#C00000}{\int_0^{1}\frac{\ln(x)}{1+x^2}dx}$$
Subsitute : $$t=\frac{1}{x}$$
$$\int_{\infty}^{1}\frac{\ln(\frac{1}{t})}{1+(\frac{1}{t})^2}\cdot-\frac{1}{t^2}dt=-\int_{\infty}^{1} \frac{\ln (t^{-1})}{t^2+1} dt=\int_{\infty}^{1} \frac{\ln (t)}{t^2+1} dt$$

Note: $$\color{blue}{\int_a^b f(t) dt= -\int_b^a f(t) dt}$$

$$\int_{\infty}^{1} \frac{\ln (t)}{1+t^2} dt=-\int_1^{\infty} \frac{\ln (t)}{1+t^2} dt$$

Back to
$$\int_0^{\infty}\frac{\ln(x)}{1+x^2}dx=\color{#C00000}{\int_0^{1}\frac{\ln(x)}{1+x^2}dx}+\int_1^{\infty} \frac{\ln(x)}{1+x^2}dx$$
$$=\color{green}{-\int_1^{\infty} \frac{\ln (t)}{1+t^2} dt+\int_1^{\infty} \frac{\ln(x)}{1+x^2}dx}$$

Im sure you  can see that $x$ and $t$ are just to letters assigned to the integral , that:
$$\int_1^{\infty} \frac{\ln (t)}{1+t^2} dt=\int_1^{\infty} \frac{\ln(x)}{1+x^2}dx$$

Therefore:
$$-\int_1^{\infty} \frac{\ln (t)}{1+t^2} dt+\int_1^{\infty} \frac{\ln(x)}{1+x^2}dx=0$$
$$\int_0^{\infty} \frac{\ln(x)}{1+x^2}dx=\color{green}{-\int_1^{\infty} \frac{\ln (t)}{1+t^2} dt+\int_1^{\infty} \frac{\ln(x)}{1+x^2}dx}=0$$
A: Hint
Just write $$I=\int_0^{\infty}\frac{\ln(x)}{1+x^2}dx=\int_0^1\frac{\ln(x)}{1+x^2}dx+\int_1^{\infty}\frac{\ln(x)}{1+x^2}dx$$ For the second integral, change variable $x=\frac{1}{y}$
