What is the integral of r^2((r^3)/9-8)^2 ?

The integral of r^2((r^3)/9-8)^2 is ((r^3)/9-8)^3+C

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integral of r^2((r^3)/9-8)^2

\int r^{2} (\frac{r ^{3}}{9} - 8)^{2} d r

(\frac{r ^{3}}{9} - 8)^{3} + C

Solution steps

\int r^{2} (\frac{r ^{3}}{9} - 8)^{2} d r

Apply u-substitution

: \int 3 u^{2} d u

= \int 3 u^{2} d u

Take the constant out:

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

= 3 \cdot \int u^{2} d u

Apply the Power Rule

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

= 3 \cdot \frac{u ^{3}}{3}

Substitute back

u = \frac{r ^{3}}{9} - 8

= 3 \cdot \frac{( \frac{r ^{3}}{9} - 8 ) ^{3}}{3}

Simplify

3 \cdot \frac{( \frac{r ^{3}}{9} - 8 ) ^{3}}{3} : (\frac{r ^{3}}{9} - 8)^{3}

= (\frac{r ^{3}}{9} - 8)^{3}

Add a constant to the solution

= (\frac{r ^{3}}{9} - 8)^{3} + C

f^{^{'}} (x) = \frac{3 f ( x )}{x ln ( x )}

\frac{d}{d x} (\frac{14400}{x})

\frac{d}{d x} y = \frac{x ^{4} - 3 x ^{3} - 6 x ^{2} - 3 x + 2}{x}

\int \frac{5}{x ^{2} + 5} d x

n = 1 \sum \infty \frac{6}{n ^{5}}

What is the integral of r^2((r^3)/9-8)^2 ?
The integral of r^2((r^3)/9-8)^2 is ((r^3)/9-8)^3+C