# 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.

• Try partial fractions decomposition on $\frac{-2x-6}{x^4-x^2-2x}$ (or on the last integrand). – kccu May 1 '17 at 12:23
• A good thing to note if the power on the numerator is one less than the power on the denominator, then you should try to make the numerator look like the derivative of the denominator by doing algebraic manipulation such as multiplying (and then dividing to cancel the factor) or adding (and then subtracting that same term) etc. – Twenty-six colours May 1 '17 at 12:38

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.