I am trying to evaluate $\int \frac{1}{x\ln(x-2)} d x + \frac{1}{(x-2)\ln(x-2)} dx$
I tried using substitution $u = x-2 \iff dx = \frac{du}{x-2}$
And ended up with
$$\int \frac{1}{x\ln(x-2)} dx + \ln(\ln(x-2)) + c_1 $$
Still I have a hard time calculating $$\int \frac{1}{x\ln(x-2)} dx $$
Any ideas?
A different approach that may work
$\int \frac{1}{x\ln(x-2)} + \frac{1}{(x-2)\ln(x-2)} dx = (-\frac{1}{2}) \int \frac{\frac{1}{x} - \frac{1}{x-2}}{ln(x-2)} dx = \int \frac{1}{x(x-2)ln(x-2)} dx$
This seems easier but still I can't evaluate it.
Ahhhh we have found your problem. The differential equation is already exact. Double check the derivatives. ${\frac{d}{dx}\left(\ln(x)-2\right) = \frac{1}{x}}$, and also ${\frac{d}{dy}\left(\frac{y}{x}+6x\right)=\frac{1}{x}}$