# How do you find the definite integral of (x^4 - 1)/( x^2 + 1) dx from -5 to -2?

${\int}_{-} {5}^{-} 2 \frac{{x}^{4} - 1}{{x}^{2} + 1} \mathrm{dx} = {\int}_{-} {5}^{-} 2 \left({x}^{2} - 1\right) \mathrm{dx} =$
${\left[{x}^{3} / 3 - x\right]}_{-} {5}^{-} 2 = - \frac{8}{3} + 2 - \left(- \frac{125}{3} + 5\right) = \frac{- 8 + 6 + 125 - 15}{3} = \frac{108}{3} = 36$