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

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

Popular >

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} - 8

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

Simplify

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

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

Add a constant to the solution

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

\int sin^{5} (x) cot (x) d x

\frac{ln ( t )}{t}

\int \frac{x ^{2} - 256}{x} d x

x = - 3, x = 4, y = e^{2 x}, y = 2 e^{- x}

\frac{d}{d x} (x^{m} a^{2} - x^{2})

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