What is the integral of \sqrt[3]{(8x^2-4)^5}x ?

The integral of \sqrt[3]{(8x^2-4)^5}x is 3/128 (8x^2-4)^{8/3}+C

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integral of \sqrt[3]{(8x^2-4)^5}x

\frac{3}{128} (8 x^{2} - 4)^{\frac{8}{3}} + C

Solution steps

Apply u-substitution

= \int \frac{( 8 u - 4 ) ^{\frac{5}{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 (8 u - 4)^{\frac{5}{3}} d u

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{v ^{\frac{5}{3}}}{8} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{8} \cdot \int v^{\frac{5}{3}} d v

Apply the Power Rule

= \frac{1}{2} \cdot \frac{1}{8} \cdot \frac{3}{8} v^{\frac{8}{3}}

Substitute back

= \frac{1}{2} \cdot \frac{1}{8} \cdot \frac{3}{8} (8 x^{2} - 4)^{\frac{8}{3}}

Simplify

\frac{1}{2} \cdot \frac{1}{8} \cdot \frac{3}{8} (8 x^{2} - 4)^{\frac{8}{3}} : \frac{3}{128} (8 x^{2} - 4)^{\frac{8}{3}}

= \frac{3}{128} (8 x^{2} - 4)^{\frac{8}{3}}

Add a constant to the solution

= \frac{3}{128} (8 x^{2} - 4)^{\frac{8}{3}} + C

What is the integral of \sqrt[3]{(8x^2-4)^5}x ?
The integral of \sqrt[3]{(8x^2-4)^5}x is 3/128 (8x^2-4)^{8/3}+C