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

Question

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

Answer 1

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

$${\rm Li}_{s+1}(x)=\int_0^x\frac{{\rm Li}_s(t)}t{\rm~d}t$$

Answer 2

You can follow the steps below.

Answer 3

we know that $$\ln(x)+C=\int\frac1x{\rm~d}x$$ also

$${\rm Li}_2(x)+C=\int\frac{\ln(1-x)}x{\rm~d}x$$

$${\rm Li}_1(x)=\ln(1-x)$$

also $${\rm Li}_{s+1}(x)=\int_0^x\frac{{\rm Li}_s(t)}t{\rm~d}t$$