What is the integral of (9x^2)/(sqrt(1-x^3)) ?

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

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integral of (9x^2)/(sqrt(1-x^3))

\int \frac{9 x ^{2}}{1 - x ^{3}} d x

- 6 1 - x^{3} + C

Solution steps

\int \frac{9 x ^{2}}{1 - x ^{3}} d x

Take the constant out:

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

= 9 \cdot \int \frac{x ^{2}}{1 - x ^{3}} d x

Apply u-substitution

= 9 \cdot \int - \frac{1}{3 u} d u

Take the constant out:

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

= 9 (- \frac{1}{3} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

= 9 (- \frac{1}{3} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 1 - x^{3}

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

Simplify

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

= - 6 1 - x^{3}

Add a constant to the solution

= - 6 1 - x^{3} + C

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