<s>[INST] How to evaluate $\int \frac{2 x^3 - 3 x^2 - 26 x + 38}{x^4 - 2 x^3 - 13 x^2 + 38 x - 12} \, dx$ 
Evaluate the integral $$\int \frac{2 x^3 - 3 x^2 - 26 x + 38}{x^4 - 2
 x^3 - 13 x^2 + 38 x - 12} \, dx$$

I tried to split the integral with partial fractions but I could not find a suitable factorization for the denominator. I think I could do it if I used complex roots but that seems very complicated. 
I wasn't too sure on how to proceed from here.
 [/INST] Well, I can take a stab at starting it:
$$
\begin{align*}
\int \frac{2 x^3 - 3 x^2 - 26 x + 38}{x^4 - 2
 x^3 - 13 x^2 + 38 x - 12} \, dx &= \int \frac{4 x^3 - 6 x^2 - 26 x + 38}{x^4 - 2
 x^3 - 13 x^2 + 38 x - 12} \, dx\\ &\qquad+ \int \frac{-2x^{3} + 3x^{2}}{x^4 - 2
 x^3 - 13 x^2 + 38 x - 12}\,dx\\
&=\ln\left|x^4 - 2
 x^3 - 13 x^2 + 38 x - 12\right|\\ &\qquad+ \int\frac{-2x^{3} + 3x^{2}}{x^4 - 2
 x^3 - 13 x^2 + 38 x - 12}\,dx.
\end{align*}
$$
I don't see what you can do from there, however.
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