Evalute $\int\frac{3x-2}{x^2-4x+5} \, dx$ Evalute $$\int\frac{3x-2}{x^2-4x+5} \, dx$$
So, I tried using $x^2-4x+5 = t$, which unfortunately didn't help much.
I also tried making the denominator $(x-2)^2 +1$ and then using $x-2 = t$, which made everything messy. Any suggestions? It looks so simple, and I bet my first approach is right.

Hint. You may write the integrand as
$$
\frac{3x-2}{x^2-4x+5}=\frac32\frac{2x-4}{x^2-4x+5}+\frac{4}{x^2-4x+5}
$$ giving

$$
\frac{3x-2}{x^2-4x+5}=\frac32\frac{(x^2-4x+5)'}{x^2-4x+5}+\frac{4}{(x-2)^2 +1}
$$

Can you take it from here?
