How does one evaluate $ \int _2^4\:\frac{x^4-x^2-4x-6}{\:x^4-x^2-2x}dx $? I tried this : $$\begin{align}&\int _2^4\:\frac{x^4-x^2-4x-6}{\:x^4-x^2-2x}dx\\&=\int _2^4\:dx\:+\:\int _2^4\:\frac{-2x-6}{x^4-x^2-2x}dx\\&=2-\int _2^4\:\frac{4x^3-2x-2-4x^3-4}{x^4-x^2-2x}dx\\&=\:2-\:\ln\left(29\right)\:-4\int _2^4\:\frac{x^3+1}{x^4-x^2-2x}dx\end{align}$$ but I'm stuck at the last integral.

Hint : $${\displaystyle\int}\dfrac{x^3+1}{x^4-x^2-2x}\,\mathrm{d}x=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{3x^2-1}{x^3-x-2}\,\mathrm{d}x-\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{1}{x}\,\mathrm{d}x$$
And use the change of variable $u=x^3-x-2$ for the first integral.
