How can I calculate $\int\frac{x-2}{-x^2+2x-5}dx$? I'm completely stuck on solving this indefinite integral:
$$\int\frac{x-2}{-x^2+2x-5}dx$$
By completing the square in the denominator and separating the original into two integrals, I get:
$$-\int\frac{x}{x^2-2x+5}dx -\int\frac{2}{(x-1)^2 + 4}dx$$
The second one is trivial, but the first one has me stuck. Whatever substitution I apply or form I put it in, I just can't figure it out. They're meant to be solved without partial integration, by the way.

Since $x^2-2x+5=(x-1)^2+2^2$, the substitution to do is $2u=x-1$, so the integral becomes
$$
\int\frac{2-x}{x^2-2x+5}\,dx=
\int\frac{1-2u}{4u^2+4}2\,du=\frac{1}{2}\int\frac{1-2u}{u^2+1}\,du
$$
No need to guess, now.
