$$I(x) = \int_0^\infty \frac{\ln (x) dx}{x^2+2x+4}$$
$$I( \alpha) = \int_0^\infty \frac{\ln (\alpha x) dx}{x^2+2x+4}$$
$$I'( \alpha) = \int_0^\infty \frac{dx}{(\alpha) (x^2+2x+4)}$$
$$I'( \alpha) = \frac{1}{(\alpha)} \int_0^\infty \frac{dx}{(x+1)^2 + (\sqrt{3})^2}$$
$$ x+1=t$$
$$I'( \alpha) = \frac{1}{(\alpha)} \int_1^\infty \frac{dt}{(t^2 + (\sqrt{3})^2} $$
$$I'( \alpha) = \frac{1}{(\alpha)(\sqrt{3})} \arctan \frac{t}{\sqrt{3}} \Big|_1^\infty $$
$$I'( \alpha) = \frac{\pi}{3\times \sqrt{3}\times\alpha}$$
I am unsure what to do next as if I integrate wrt. alpha I get my limits from 1 to infinity and I am getting infinity as the resultant. The answer given is $\frac{\pi\ln2}{3\sqrt{3}}$
Best Answer
Under $x\to 2x$, one has $$I = \int_0^\infty \frac{\ln (x) dx}{x^2+2x+4} =\frac12 \int_0^\infty \frac{\ln (2x)dx}{x^2+x+1}=\frac12 \int_0^\infty \frac{\ln 2dx}{x^2+x+1}+ \frac12 \int_0^\infty \frac{\ln (x) dx}{x^2+x+1}.$$ The second integral is zero under $x\to1/x$. It is easy to get the first integral. I omit the details.