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

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

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

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

\frac{1}{4} (x^{2} - 8)^{4} + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int (x^{2} - 8)^{3} x d x

Apply u-substitution

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

= 2 \cdot \int \frac{u ^{3}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{4}}{4}

Substitute back

u = x^{2} - 8

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 8 ) ^{4}}{4}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 8 ) ^{4}}{4} : \frac{1}{4} (x^{2} - 8)^{4}

= \frac{1}{4} (x^{2} - 8)^{4}

Add a constant to the solution

= \frac{1}{4} (x^{2} - 8)^{4} + C

\int_{1}^{\infty} x^{- 7} d x

\frac{d}{d x} (\frac{x + 2}{x ^{2} + 4 x + 13})

n = 1 \sum \infty \frac{60}{7 ^{n}}

f (t) = sin (t) + t cos (t)

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

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