What is the integral of (x^2-1)/((x^3-3x)^3) ?

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

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integral of (x^2-1)/((x^3-3x)^3)

\int \frac{x ^{2} - 1}{( x ^{3} - 3 x ) ^{3}} d x

- \frac{1}{6 ( x ^{3} - 3 x ) ^{2}} + C

Solution steps

\int \frac{x ^{2} - 1}{( x ^{3} - 3 x ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{3 u ^{3}} d u

= \int \frac{1}{3 u ^{3}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= \frac{1}{3} (- \frac{1}{2 u ^{2}})

Substitute back

u = x^{3} - 3 x

= \frac{1}{3} (- \frac{1}{2 ( x ^{3} - 3 x ) ^{2}})

Simplify

\frac{1}{3} (- \frac{1}{2 ( x ^{3} - 3 x ) ^{2}}) : - \frac{1}{6 ( x ^{3} - 3 x ) ^{2}}

= - \frac{1}{6 ( x ^{3} - 3 x ) ^{2}}

Add a constant to the solution

= - \frac{1}{6 ( x ^{3} - 3 x ) ^{2}} + C

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