# 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.

• Before attempting the problem, have you tried Wolfram Alpha? It should get you an idea of how the solution should look like. – John Smith Mar 28 '17 at 0:53
• Deepak is correct. – John Smith Mar 28 '17 at 0:54
• @i8Σπ_821 I did it in Wolfram alpha but I don't really understand it because it came out as a really complicated result but it came out from my teacher's practice exercises so I don't believe it should be that complicated – John Rawls Mar 28 '17 at 0:55
• Before you changed it, the denominator in the title had an easy factorisation. Are you sure that wasn't the correct one? – Paul Castle Mar 28 '17 at 0:55
• @PaulWright that was a previous problem that I solved easily – John Rawls Mar 28 '17 at 0:56

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.

• @John, Could u complete the answer? If yes, can u please share the same? – lab bhattacharjee Mar 28 '17 at 1:32

It's strange that the exercises would have the easier question followed by the exact same question with one number changed to make it very difficult.

Either the writer made a mistake, or they are making a point about how slight changes to an integral can make them much more difficult.

If this is an assignment, I would say "let $r_1, r_2, r_3, r_4$ be the roots of the denominator" and continue the question from there.

• "Either the writer made a mistake, or they are making a point about how slight changes to an integral can make them much more difficult." Agreed. This questions is hardly instructive in its supposed solution but more instructive about demonstrating its difficulty. – Xoque55 Mar 28 '17 at 1:24