What is the integral of r^2((r^3)/(18)-1)^5 ?

The integral of r^2((r^3)/(18)-1)^5 is ((r^3)/(18)-1)^6+C

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integral of r^2((r^3)/(18)-1)^5

\int r^{2} (\frac{r ^{3}}{18} - 1)^{5} d r

(\frac{r ^{3}}{18} - 1)^{6} + C

Solution steps

\int r^{2} (\frac{r ^{3}}{18} - 1)^{5} d r

Apply u-substitution

: \int 6 u^{5} d u

= \int 6 u^{5} d u

Take the constant out:

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

= 6 \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 6 \cdot \frac{u ^{6}}{6}

Substitute back

u = \frac{r ^{3}}{18} - 1

= 6 \cdot \frac{( \frac{r ^{3}}{18} - 1 ) ^{6}}{6}

Simplify

6 \cdot \frac{( \frac{r ^{3}}{18} - 1 ) ^{6}}{6} : (\frac{r ^{3}}{18} - 1)^{6}

= (\frac{r ^{3}}{18} - 1)^{6}

Add a constant to the solution

= (\frac{r ^{3}}{18} - 1)^{6} + C

\int cos (x) ln (x) sin (x) d x

\int \frac{2 ^{x}}{2 ^{x} + 3} d x

\frac{d}{d x} (2 2 x)

n = 0 \sum \infty \frac{5}{n + 2}

y = x^{4}

What is the integral of r^2((r^3)/(18)-1)^5 ?
The integral of r^2((r^3)/(18)-1)^5 is ((r^3)/(18)-1)^6+C