<s>[INST] 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.
 [/INST] 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?
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