Indefinite integral of $\frac{\ln(x)}{(x-3)^2}$ I am trying to compute the integral
$$
\int\frac{\ln(x)}{(x-3)^2}\mathrm{d}x
$$
I have tried the following substitution, but seem to get nowhere: $u = x - 3$.
$$x=u+3$$
$$dx=du$$
$$\int\frac{\ln(u+3)}{u^3}\mathrm{d}u$$
I get stuck at this point and do not know what to do, could anyone lead me in the right direction?
 A: Note that $u$ is squared, not cubed, in the denominator.
$$I = \int\frac{\ln(u+3)}{u^3}\mathrm{d}u$$
Using integration by parts, take $$w = \ln(u+3) \implies dw= \frac{1}{u+3}\,du$$ Then $$dv = \frac 1{u^2} \implies v = -\frac{1}{u}$$
$$I = wv - \int v\,dw$$
$$I = -\frac{\ln(u + 3)}{u} + \int \frac 1{u}\cdot \frac 1{u+3}\,du$$
Now you can use partial fraction decomposition $$\int \frac
   1{u(u+3)}\,du =\int\left(\frac Au + \frac B{u+1}\right)\,du$$
Or, you can complete the square in the denominator of the remaining
   integral to use trig substitution. $$u(u+3) = u^2 + 3u + \frac 94
   -\frac 94 = \left(u + \frac 32\right)^2 - \left(\frac 32\right)^2$$
Now put $\sec \theta = \frac 32\left(u + \frac 32\right)$, and use the fact that $$\sec^2 \theta -1 = \tan^2\theta$$
A: Let $u=\ln x$ and $dv=\frac{1}{(x-3)^2} dx$, so $du=\frac{1}{x} dx$ and $v=-\frac{1}{x-3}$.
Then $$\displaystyle\int\frac{\ln x}{(x-3)^2} \, dx=uv-\int v\,du=-\frac{\ln x}{x-3}-\int-\frac{1}{x(x-3)}\, dx=-\frac{\ln x}{x-3}+\int\frac{1}{x(x-3)}\, dx.$$
Now use partial fractions: 
$$\frac{1}{x(x-3)}=\frac{A}{x}+\frac{B}{x-3}\implies A(x-3)+Bx=1\implies A=-\frac{1}{3} \text{ and } B=\frac{1}{3}.$$
