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

The integral of r^2(r^3+6)^8 is 1/27 (r^3+6)^9+C

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

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

\frac{1}{27} (r^{3} + 6)^{9} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{8}}{3} d u

= \int \frac{u ^{8}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u^{8} d u

Apply the Power Rule

: \frac{u ^{9}}{9}

= \frac{1}{3} \cdot \frac{u ^{9}}{9}

Substitute back

u = r^{3} + 6

= \frac{1}{3} \cdot \frac{( r ^{3} + 6 ) ^{9}}{9}

Simplify

\frac{1}{3} \cdot \frac{( r ^{3} + 6 ) ^{9}}{9} : \frac{1}{27} (r^{3} + 6)^{9}

= \frac{1}{27} (r^{3} + 6)^{9}

Add a constant to the solution

= \frac{1}{27} (r^{3} + 6)^{9} + C

(x + 1) (x - 2) y^{^{'}} + y = 0

\frac{d y}{d x} - \frac{1}{x} y = x y^{2}

\int (3 - 2 x)^{10} d x

\int_{1}^{2} \frac{x ^{2} + 4}{3 x - x ^{2}} d x

\frac{\partial}{\partial y} (\frac{y}{y - x})

What is the integral of r^2(r^3+6)^8 ?
The integral of r^2(r^3+6)^8 is 1/27 (r^3+6)^9+C