# How do you find the indefinite integral of ∫ x^2 - 2 dx / x^3 - 4x ?

Apr 15, 2018

$I = \frac{1}{4} \ln \left({x}^{4} - 4 {x}^{2}\right) + C$

#### Explanation:

We want to solve

$I = \int \frac{{x}^{2} - 2}{{x}^{3} - 4 x} \mathrm{dx}$

Multiply the DEN and NUM by $x$

$I = \int \frac{{x}^{3} - 2 x}{{x}^{4} - 4 {x}^{2}} \mathrm{dx}$

Now we can make i nice substitution
color(red)(u=x^4-4x^2=>du=4x^3-8xdx=4(x^3-2x)dx

$I = \frac{1}{4} \int \frac{1}{u} \mathrm{du}$

$\textcolor{w h i t e}{I} = \frac{1}{4} \ln \left(u\right) + C$

$\textcolor{w h i t e}{I} = \frac{1}{4} \ln \left({x}^{4} - 4 {x}^{2}\right) + C$

I have solved this way, applying partial fractions decomposition: