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.

Another plan that may be useful: once we see that form with the completed square, we make a simple substitution - not the whole thing, but just the part inside the square.
\begin{align*}I &= -\int \frac{x-2}{(x-1)^2+4}\,dx\\
&\phantom{|}^{u=x-1}_{du=dx}\\
&= \int -\frac{u-1}{u^2+4}\,du = \int\frac{-u}{u^2+4}\,du+\int\frac{1}{u^2+4}\,du\end{align*}
The substance of this is exactly the same as @E-mu's argument - the difference is how we arrive at the way to split the integrand. Instead of looking for the derivative of the denominator, we make an affine substitution so that the split becomes obvious.
