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

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

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

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

- 9 - 4 x^{3} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{12 u} d u

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

Take the constant out:

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

= 6 (- \frac{1}{12} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

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

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

Substitute back

u = 9 - 4 x^{3}

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

Simplify

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

= - 9 - 4 x^{3}

Add a constant to the solution

= - 9 - 4 x^{3} + C

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