$$\int_4^8 \frac {\ln(9-x)}{\ln(9-x)+\ln(x-3)}\,dx$$
So far I've tried using u substitution with $u = 12-x$, which is what my teacher recommended, and put that equation in terms of $x = 12-u$ and plugged that in, but I got stuck there. Any and all help is appreciated!
$$I=\int_4^8 \frac{\ln(9-x)}{\ln(9-x) + \ln(x-3)}$$
$u=12-x \implies du = -dx$
$$I=-\int_8^4 \frac{\ln(u-3)}{\ln(u-3) + \ln(9-u)} du = \int_4^8 \frac{\ln(x-3)}{\ln(x-3) + \ln(9-x)}$$
Adding both expressions:
$$2I = \int_4^8 1 dx = 4 \implies I=2.$$
Edit: Note that $u$ and $x$ are dummy variables (for example, is the same writing $f(x) =x$ or $f(t)=t$).
