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

The integral of x^2(x^3+8)^2 is 1/9 (x^3+8)^3+C

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integral of x^2(x^3+8)^2

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

\frac{1}{9} (x^{3} + 8)^{3} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2}}{3} d u

= \int \frac{u ^{2}}{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^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

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

Substitute back

u = x^{3} + 8

= \frac{1}{3} \cdot \frac{( x ^{3} + 8 ) ^{3}}{3}

Simplify

\frac{1}{3} \cdot \frac{( x ^{3} + 8 ) ^{3}}{3} : \frac{1}{9} (x^{3} + 8)^{3}

= \frac{1}{9} (x^{3} + 8)^{3}

Add a constant to the solution

= \frac{1}{9} (x^{3} + 8)^{3} + C

n = 1 \sum \infty 2 (0.5)^{n - 1}

\frac{\partial}{\partial x} (ln (\frac{1}{x ^{2} + y ^{2}}))

\frac{d}{d x} (\frac{x ^{2} - 2}{x ^{2} + 2})

\frac{\partial}{\partial x} (6 e^{x^{3} y^{2}})

5 x^{2} + 5 x + 2

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