What is the integral of r^2((r^3)/(15)-8)^4 ?

The integral of r^2((r^3)/(15)-8)^4 is ((r^3)/(15)-8)^5+C

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integral of r^2((r^3)/(15)-8)^4

\int r^{2} (\frac{r ^{3}}{15} - 8)^{4} d r

(\frac{r ^{3}}{15} - 8)^{5} + C

Solution steps

\int r^{2} (\frac{r ^{3}}{15} - 8)^{4} d r

Apply u-substitution

: \int 5 u^{4} d u

= \int 5 u^{4} d u

Take the constant out:

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

= 5 \cdot \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

= 5 \cdot \frac{u ^{5}}{5}

Substitute back

u = \frac{r ^{3}}{15} - 8

= 5 \cdot \frac{( \frac{r ^{3}}{15} - 8 ) ^{5}}{5}

Simplify

5 \cdot \frac{( \frac{r ^{3}}{15} - 8 ) ^{5}}{5} : (\frac{r ^{3}}{15} - 8)^{5}

= (\frac{r ^{3}}{15} - 8)^{5}

Add a constant to the solution

= (\frac{r ^{3}}{15} - 8)^{5} + C

\frac{\partial}{\partial x} (\frac{s ^{2} + 4}{t - 3})

\frac{d}{d x} (ln (arctan (2 x^{5})))

\int \frac{1}{6 x ^{5}} d x

3 \cdot 2^{2 x + 1}

\int_{0}^{1} r 1 - r^{2} d r

What is the integral of r^2((r^3)/(15)-8)^4 ?
The integral of r^2((r^3)/(15)-8)^4 is ((r^3)/(15)-8)^5+C