What is the integral of r^2(r^3-4)^7 ?

The integral of r^2(r^3-4)^7 is 1/24 (r^3-4)^8+C

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integral of r^2(r^3-4)^7

\int r^{2} (r^{3} - 4)^{7} d r

\frac{1}{24} (r^{3} - 4)^{8} + C

Solution steps

\int r^{2} (r^{3} - 4)^{7} d r

Apply u-substitution

= \int \frac{u ^{7}}{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^{7} d u

Apply the Power Rule

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

Substitute back

u = r^{3} - 4

= \frac{1}{3} \cdot \frac{( r ^{3} - 4 ) ^{8}}{8}

Simplify

\frac{1}{3} \cdot \frac{( r ^{3} - 4 ) ^{8}}{8} : \frac{1}{24} (r^{3} - 4)^{8}

= \frac{1}{24} (r^{3} - 4)^{8}

Add a constant to the solution

= \frac{1}{24} (r^{3} - 4)^{8} + C

\frac{d y}{d x} + 2 y = 2 x

y^{^{'}} + \frac{1}{x} y = x y^{2}

x y + x

f (x) = \frac{6}{x ^{2}}, (1, 6)

\frac{d}{d x} ((f (x) (3 x))^{2} - 4)

What is the integral of r^2(r^3-4)^7 ?
The integral of r^2(r^3-4)^7 is 1/24 (r^3-4)^8+C