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.
 A: 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?
A: 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 $u = x-2$.
A: Observe that $$\frac{3x-2}{x^2-4x+5}=\frac{3(x-2)}{(x-2)^2+1}+\frac{4}{(x-2)^2+1}$$
The integral of the first summand can be computed using your first substitution and the second using your second substitution.
A: Using your second idea, you get 
$\displaystyle\int\frac{3(t+2)-2}{t^2+1}\;dt=\int\frac{3t+4}{t^2+1}\;dt=\frac{3}{2}\int\frac{2t}{t^2+1}\;dt+4\int\frac{1}{t^2+1}\;dt$
$\displaystyle=\frac{3}{2}\ln(t^2+1)+4\arctan t+C=\frac{3}{2}\ln(x^2-4x+5)+4\arctan(x-2)+C$
A: Try partial fraction decomposition
