finding $ \int^{\infty}_{0}\frac{\ln x}{x^2+6x+9}dx$ 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
 A: Hint. Alternatively integrating by parts, with $M>0$ and $\varepsilon>0$, one gets
$$
\int^M_{\varepsilon}\frac{\ln x}{(x+3)^2}dx=\left[\left(\frac13-\frac{1}{(x+3)}\right)\cdot \ln x\right]^M_{\varepsilon}-\int^M_{\varepsilon}\left(\frac13-\frac{1}{(x+3)}\right)\cdot \frac1x\:dx,
$$ the latter integral is easier, then one may let $M \to \infty$, $\varepsilon \to 0^+$.
A: Hint. Try integration by parts. Take 
$$u=\ln x \qquad \text{and} \qquad dv=\frac{dx}{(x+3)^2}.$$
Then, the integral is equal to
$$\int_{\frac{1}{M}}^M \frac{\ln x}{(x+3)^2}dx=\left[-\frac{\ln x}{x+3}\right]_{\frac{1}{M}}^M+\int_{\frac{1}{M}}^M \frac{dx}{(x+3)x}$$
You can solve the last integral applying the theory of rational integrals and take the limit $M \to \infty$.
A: 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 $\boxed{\dfrac{\ln 3}{3}}$
