What is the integral of (4x^3-12x)(x^4-6x^2)^{-3} ?

The integral of (4x^3-12x)(x^4-6x^2)^{-3} is -1/(2(x^4-6x^2)^2)+C

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integral of (4x^3-12x)(x^4-6x^2)^{-3}

\int (4 x^{3} - 12 x) (x^{4} - 6 x^{2})^{- 3} d x

- \frac{1}{2 ( x ^{4} - 6 x ^{2} ) ^{2}} + C

Solution steps

\int (4 x^{3} - 12 x) (x^{4} - 6 x^{2})^{- 3} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

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

= - \frac{1}{2 ( x ^{4} - 6 x ^{2} ) ^{2}}

Add a constant to the solution

= - \frac{1}{2 ( x ^{4} - 6 x ^{2} ) ^{2}} + C

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What is the integral of (4x^3-12x)(x^4-6x^2)^{-3} ?
The integral of (4x^3-12x)(x^4-6x^2)^{-3} is -1/(2(x^4-6x^2)^2)+C