Compute a Integral Probably Applying Residue

Question

I have been trying to compute $$\displaystyle \int_{0}^{+\infty} \dfrac{\ln x}{\left( 2 x^2 - x + 6 \right)^2} dx$$. I consider it by using Residue Theorem. Considering a complex function: $$f \left( z \right) = \dfrac{\ln^2 z}{\left( 2 z^2 - z + 6 \right)^2}$$After taking a look at $f(z)$, there are two singularity points both of order 2, but they keeps quite nasty forms as a unfriendly pair of roots of quadratic equation. With great struggle, I found the sum of the residue is $$\left( \dfrac{8 \ln 3}{47 \sqrt{47}} - \dfrac{\sqrt{47}}{141} \right) \arctan \sqrt{47} - \frac{\ln 3}{282}$$I cannot proceed to figure out the result and I am hesitating whether there is a better way to compute it.

Answer 1

Without residues.

Call $a$ and $b$ the roots of $2x^2-x+6=0$. So $$\dfrac{1}{\left( 2 x^2 - x + 6 \right)^2}= \frac{1}{4 (x-a)^2 (x-b)^2}$$ Using partial fraction decomposition $$\frac{1}{(x-a)^2 (x-b)^2}=\frac{2}{(a-b)^3 (x-b)}+\frac{1}{(a-b)^2 (x-b)^2}-\frac{2}{(a-b)^3 (x-a)}+\frac{1}{(a-b)^2 (x-a)^2}$$ making that you face things such as $$\int \frac {\log(x)}{x-k}\,dx=\text{Li}_2\left(\frac{x}{k}\right)+\log (x) \log \left(1-\frac{x}{k}\right)$$ $$\int \frac {\log(x)}{(x-k)^2}\,dx=-\frac{\log (x)}{k}-\frac{\log (x)}{x-k}+\frac{\log (x-k)}{k}$$