Mathematical Reflections 2016, Issue 1 - Problem U364

Problem:
Evaluate $$\int \dfrac{5x^2-x-4}{x^5+x^4+1}\ dx.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Observe that $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \dfrac{5x^2-x-4}{x^5+x^4+1}&=&\dfrac{5x^2-x-4}{(x^3-x+1)(x^2+x+1)}\\&=&-\dfrac{3(x+1)}{x^2+x+1}+\dfrac{3x^2-1}{x^3-x+1}\\&=&-\dfrac{3}{2}\cdot\dfrac{2x+1}{x^2+x+1}+\dfrac{3x^2-1}{x^3-x+1}-\dfrac{3}{2}\cdot\dfrac{1}{x^2+x+1} \\ &=&-\dfrac{3}{2}\cdot\dfrac{2x+1}{x^2+x+1}+\dfrac{3x^2-1}{x^3-x+1}-\sqrt{3}\cdot\dfrac{2/\sqrt{3}}{\left(\frac{2x+1}{\sqrt{3}}\right)^2+1}. \end{array}$$
So, $$\renewcommand{\arraystretch}{2} \begin{array}{lll} \displaystyle \int \dfrac{5x^2-x-4}{x^5+x^4+1}\ dx&=&\displaystyle -\dfrac{3}{2}\int \dfrac{2x+1}{x^2+x+1} \ dx+\int \dfrac{3x^2-1}{x^3-x+1} \ dx-\sqrt{3}\int \dfrac{2/\sqrt{3}}{\left(\frac{2x+1}{\sqrt{3}}\right)^2+1} \ dx \\ &=&-\dfrac{3}{2}\log(x^2+x+1)+\log|x^3-x+1|-\sqrt{3}\arctan \dfrac{2x+1}{\sqrt{3}}+C. \end{array}$$