TITLE: finding $ \int^{\infty}_{0}\frac{\ln x}{x^2+6x+9}dx$
QUESTION [5 upvotes]: finding $\displaystyle \int^{\infty}_{0}\frac{\ln x}{x^2+6x+9}dx$ 
Attempt: let $\displaystyle I(a) = \int^{\infty}_{0}\frac{\ln (ax)}{(x+3)^2}dx, a>0$
$\displaystyle I'(a) = \int^{\infty}_{0}\frac{x}{ax(x+3)^2}dx = \frac{1}{a}\int^{\infty}_{0}\frac{1}{(x+3)^2}dx = -\frac{1}{a}\bigg(\frac{1}{x+3}\bigg)\bigg|_{0}^{\infty} = \frac{1}{3a}$
so $\displaystyle I(a) = \frac{\ln(a)}{3}+C$
could some help me how  to solve from there, thanks in advanced

REPLY [2 votes]: If  $\displaystyle I=\int\limits_0^\infty \dfrac{\ln x}{(x+a)^2}\; dx$ for some $a\gt 0$,
Then the substitution $xy=a^2$ gives 
$\displaystyle I=\int\limits_0^\infty \dfrac{2\ln(a)-\ln(y)}{(y+a)^2}\; dy\\2I=\displaystyle 2\ln a\int\limits_0^\infty \dfrac{dy}{(y+a)^2} = \dfrac{2\ln a}{a}$
Therefore, $\displaystyle \int\limits_0^\infty \dfrac{\ln x}{(x+a)^2}\; dx = \dfrac{\ln a}{a}$
For this one, the answer is $\boxed{\dfrac{\ln 3}{3}}$