How can I calculate $\int\frac{x-2}{-x^2+2x-5}dx$?

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?

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.