Answer
$$x{\log _5}x - \frac{x}{{\ln 5}} + C{\text{ }}$$
Work Step by Step
$$\eqalign{
& \int {{{\log }_5}x} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = {\log _5}x,{\text{ }}du = \frac{1}{{x\ln 5}}dx{\text{ }} \cr
& dv = dx,{\text{ }}v = x \cr
& \int u dv = uv - \int v du \cr
& \int {{{\log }_5}x} dx = x{\log _5}x - \int {x\left( {\frac{1}{{x\ln 5}}} \right)} dx{\text{ }} \cr
& \int {{{\log }_5}x} dx = x{\log _5}x - \int {\left( {\frac{1}{{\ln 5}}} \right)} dx{\text{ }} \cr
& \int {{{\log }_5}x} dx = x{\log _5}x - \frac{x}{{\ln 5}} + C{\text{ }} \cr} $$
