Workaround for the definite integral $\int_1^2 x \ln\left(\frac1{x^2+1}\right)dx$

Question

$$\int_1^2 x \ln\left(\frac1{x^2+1}\right)\,dx$$ I asked about the indefinite form of this one a while ago and one of the comments got me thinking. Is there a way to calculate this definite integral "directly", without wrestling with the polygarithm? I noticed it's an odd function, thus $\int_{-a}^af(x)dx=0$, but I'm not sure how to follow up on that.

Answer 1

Apparently this is all too easy. $$\int x\ln\frac1{x^2+1}\,dx$$ $$=-\frac12\int2x\ln(x^2+1)\,dx$$ $$=-\frac12(x^2+1)(\ln(x^2+1)-1)+K$$ where we have used $\int\ln x\,dx=x(\ln x-1)+K$.

Answer 2

Hint. Note that by integration by parts $$\int_1^2 x \ln\left(\frac1{x^2+1}\right)\,dx=-\int_1^2 x\ln(x^2+1)\,dx= -\left[\frac{(x^2+1)}{2}\cdot\ln(x^2+1)\right]_1^2+\int_1^2 x\,dx.$$ P.S. On the other hand you will need the polygarithm for $\int_1^2 \frac{1}{x}\ln(x^2+1)\,dx$.