<s>[INST] 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.
 [/INST] Using your second idea, you get 
$\displaystyle\int\frac{3(t+2)-2}{t^2+1}\;dt=\int\frac{3t+4}{t^2+1}\;dt=\frac{3}{2}\int\frac{2t}{t^2+1}\;dt+4\int\frac{1}{t^2+1}\;dt$
$\displaystyle=\frac{3}{2}\ln(t^2+1)+4\arctan t+C=\frac{3}{2}\ln(x^2-4x+5)+4\arctan(x-2)+C$
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