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

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

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

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

- \frac{1}{4 ( 4 x ^{3} - 9 ) ^{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{12 u ^{3}} d u

= 6 \cdot \int \frac{1}{12 u ^{3}} d u

Take the constant out:

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

= 6 \cdot \frac{1}{12} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 6 \cdot \frac{1}{12} (- \frac{1}{2 u ^{2}})

Substitute back

u = 4 x^{3} - 9

= 6 \cdot \frac{1}{12} (- \frac{1}{2 ( 4 x ^{3} - 9 ) ^{2}})

Simplify

6 \cdot \frac{1}{12} (- \frac{1}{2 ( 4 x ^{3} - 9 ) ^{2}}) : - \frac{1}{4 ( 4 x ^{3} - 9 ) ^{2}}

= - \frac{1}{4 ( 4 x ^{3} - 9 ) ^{2}}

Add a constant to the solution

= - \frac{1}{4 ( 4 x ^{3} - 9 ) ^{2}} + C

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