Integration question on $\int \frac{x}{x^2-10x+50} \, dx$ How would I integrate
$$\int \frac{x}{x^2-10x+50} \, dx$$
I am not sure on how to start the problem

write the integral as 
$$\int\frac{x}{x^2 - 10x +50}dx = \int \left(\frac{2x-10}{2(x^2-10x+50)} + \frac{5}{x^2-10x+50}\right)dx$$
Let $u = x^2 -10x +50, \; \text{then } du = 2x-10$
$$\frac{1}{2}\int\frac{1}{u}du + 5\int \frac{1}{(x-5)^2 + 25}dx$$
$$\frac{\log (u)}{2}+5\int\frac{1}{\frac{(x-5)^2}{25}+1}dx$$
$$\frac{1}{2}\log(x^2-10x+50) + \arctan\left(\frac{x-5}{5}\right) + C$$
