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

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

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

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

- \frac{1}{3 ( 3 x ^{2} - 4 ) ^{3}} + C

Solution steps

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

Take the constant out:

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

= 6 \cdot \int \frac{x}{( 3 x ^{2} - 4 ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{6 u ^{4}} d u

= 6 \cdot \int \frac{1}{6 u ^{4}} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= 6 \cdot \frac{1}{6} (- \frac{1}{3 u ^{3}})

Substitute back

u = 3 x^{2} - 4

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

Simplify

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

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

Add a constant to the solution

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

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