Answer
$$\int \frac{\ln(x-5)}{x-5}dx=\frac{\Big(\ln(x-5)\Big)^2}{2}+C$$
Work Step by Step
$$A=\int \frac{\ln(x-5)}{x-5}dx$$
Set $u=\ln(x-5)$, we have $$du=\frac{(x-5)'}{x-5}dx=\frac{1}{x-5}dx$$
Therefore, $$A=\int udu=\frac{u^2}{2}+C$$ $$A=\frac{\Big(\ln(x-5)\Big)^2}{2}+C$$
