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.

The numerator of the integrand almost looks like the derivative of the denominator, which indicates that the solution might involve a logarithm. Let's follow our nose and see if we can arrive at such an expression (you have the right idea of splitting the integral!):
\begin{align}\int\frac{x-2}{-x^2+2x-5}dx &= - \int \frac{x-2}{x^2 - 2x + 5}dx
\\&= -\frac 12 \int \frac{2x-4}{x^2 - 2x + 5}dx. \end{align} The numerator of the integrand is looking more like the derivative of the denominator, but not quite. Further manipulation yields
\begin{align}
 \int\frac{x-2}{-x^2+2x-5}dx = -\frac 12 \left(\int \frac{2x-2}{x^2-2x+5}dx  - \int \frac{2}{x^2-2x+5}dx\right).
\end{align}
The second integral you can solve, can you solve the first?
