<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] The denominator is an irreducible quadratic.  First thing you should do is write $$3x-2 = \tfrac{3}{2}(2x - \tfrac{4}{3}) = \tfrac{3}{2}(2x - 4) + 4,$$ so that the integrand becomes $$\frac{3}{2}\int \frac{2x-4}{x^2 - 4x + 5} \, dx + 4 \int \frac{dx}{x^2-4x+5}.$$  The first integral is trivial.  The second readily yields to a substitution of the form you tried; i.e., $$x^2 - 4x + 5 = (x-2)^2 + 1,$$ and choose $u = x-2$.
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