What is the integral of (6x^2-8)/((x^3-4x-1)^7) ?

The integral of (6x^2-8)/((x^3-4x-1)^7) is -1/(3(x^3-4x-1)^6)+C

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integral of (6x^2-8)/((x^3-4x-1)^7)

\int \frac{6 x ^{2} - 8}{( x ^{3} - 4 x - 1 ) ^{7}} d x

- \frac{1}{3 ( x ^{3} - 4 x - 1 ) ^{6}} + C

Solution steps

\int \frac{6 x ^{2} - 8}{( x ^{3} - 4 x - 1 ) ^{7}} d x

Apply u-substitution

: \int \frac{2}{u ^{7}} d u

= \int \frac{2}{u ^{7}} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{u ^{7}} d u

Apply the Power Rule

: - \frac{1}{6 u ^{6}}

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

Substitute back

u = x^{3} - 4 x - 1

= 2 (- \frac{1}{6 ( x ^{3} - 4 x - 1 ) ^{6}})

Simplify

2 (- \frac{1}{6 ( x ^{3} - 4 x - 1 ) ^{6}}) : - \frac{1}{3 ( x ^{3} - 4 x - 1 ) ^{6}}

= - \frac{1}{3 ( x ^{3} - 4 x - 1 ) ^{6}}

Add a constant to the solution

= - \frac{1}{3 ( x ^{3} - 4 x - 1 ) ^{6}} + C

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What is the integral of (6x^2-8)/((x^3-4x-1)^7) ?
The integral of (6x^2-8)/((x^3-4x-1)^7) is -1/(3(x^3-4x-1)^6)+C