Integral $\int_1^3 \frac{\log x}{x^2+3}dx$

Question

I stumbled upon this integral $$I=\int_1^3 \frac{\log x}{x^2+3}dx$$ Since the answer given by Wolfram-Alpha is $I=\frac{\pi \log 3}{12\sqrt3}$ I tried to work near the bounds of the integral because as a indefinite integral it will involve some polylogarythms which I want to avoid. But not a single substitution like $x=\frac{1}{y}$ or $x-2=u$ gave any better look to this integral. And when integrated by parts I feel like it got worse. Could you perhaps help me on this?

Answer 1

Hint One of your attempts is very close to one that gives a clean solution.

Substituting $x = \frac{3}{u}, dx = -\frac{3}{u^2} du$ gives $$\int_1^3 \frac{\log x\, dx}{x^2 + 3} = \int_3^1 \left(\frac{\log u - \log 3}{u^2 + 3}\right) du = -\int_1^3 \frac{\log u\, du}{u^2 + 3} + \log 3 \int_1^3 \frac{du}{u^2 + 3} .$$

The second term on the right is a standard integral. What do you notice about the first term on the right?