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

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Answer 1 · 2018-04-15T09:38:40

$I=1/4ln(x^4-4x^2)+C$

We want to solve

$I=int(x^2-2)/(x^3-4x)dx$

Multiply the DEN and NUM by $x$

$I=int(x^3-2x)/(x^4-4x^2)dx$

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

$I=1/4int1/udu$

$color(white)(I)=1/4ln(u)+C$

$color(white)(I)=1/4ln(x^4-4x^2)+C$

Answer 2 · 2018-04-16T04:08:05

I have solved this way, applying partial fractions decomposition:
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