What is the integral of (4x^3-4x)/(x^4-2x^2+5) ?

The integral of (4x^3-4x)/(x^4-2x^2+5) is ln|x^4-2x^2+5|+C

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integral of (4x^3-4x)/(x^4-2x^2+5)

\int \frac{4 x ^{3} - 4 x}{x ^{4} - 2 x ^{2} + 5} d x

ln x^{4} - 2 x^{2} + 5 + C

Solution steps

\int \frac{4 x ^{3} - 4 x}{x ^{4} - 2 x ^{2} + 5} d x

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{4} - 2 x^{2} + 5

= ln x^{4} - 2 x^{2} + 5

Add a constant to the solution

= ln x^{4} - 2 x^{2} + 5 + C

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What is the integral of (4x^3-4x)/(x^4-2x^2+5) ?
The integral of (4x^3-4x)/(x^4-2x^2+5) is ln|x^4-2x^2+5|+C