What is the integral of (x^3-3x)/(x^4-6x^2-9) ?

The integral of (x^3-3x)/(x^4-6x^2-9) is 1/4 ln|x^4-6x^2-9|+C

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integral of (x^3-3x)/(x^4-6x^2-9)

\int \frac{x ^{3} - 3 x}{x ^{4} - 6 x ^{2} - 9} d x

\frac{1}{4} ln x^{4} - 6 x^{2} - 9 + C

Solution steps

\int \frac{x ^{3} - 3 x}{x ^{4} - 6 x ^{2} - 9} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} - 6 x^{2} - 9

= \frac{1}{4} ln x^{4} - 6 x^{2} - 9

Add a constant to the solution

= \frac{1}{4} ln x^{4} - 6 x^{2} - 9 + C

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What is the integral of (x^3-3x)/(x^4-6x^2-9) ?
The integral of (x^3-3x)/(x^4-6x^2-9) is 1/4 ln|x^4-6x^2-9|+C