What is the integral of r^2((r^3)/(21)+8)^6 ?

The integral of r^2((r^3)/(21)+8)^6 is ((r^3)/(21)+8)^7+C

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integral of r^2((r^3)/(21)+8)^6

\int r^{2} (\frac{r ^{3}}{21} + 8)^{6} d r

(\frac{r ^{3}}{21} + 8)^{7} + C

Solution steps

\int r^{2} (\frac{r ^{3}}{21} + 8)^{6} d r

Apply u-substitution

: \int 7 u^{6} d u

= \int 7 u^{6} d u

Take the constant out:

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

= 7 \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= 7 \cdot \frac{u ^{7}}{7}

Substitute back

u = \frac{r ^{3}}{21} + 8

= 7 \cdot \frac{( \frac{r ^{3}}{21} + 8 ) ^{7}}{7}

Simplify

7 \cdot \frac{( \frac{r ^{3}}{21} + 8 ) ^{7}}{7} : (\frac{r ^{3}}{21} + 8)^{7}

= (\frac{r ^{3}}{21} + 8)^{7}

Add a constant to the solution

= (\frac{r ^{3}}{21} + 8)^{7} + C

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What is the integral of r^2((r^3)/(21)+8)^6 ?
The integral of r^2((r^3)/(21)+8)^6 is ((r^3)/(21)+8)^7+C