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

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

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

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

- 6 1 - r^{3} + C

Solution steps

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

Take the constant out:

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

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

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 - r^{3}

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

Simplify

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

= - 6 1 - r^{3}

Add a constant to the solution

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

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