What is the integral of 1/2 x^2\sqrt[5]{x^3-8} ?

The integral of 1/2 x^2\sqrt[5]{x^3-8} is 5/36 (x^3-8)^{6/5}+C

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integral of 1/2 x^2\sqrt[5]{x^3-8}

\frac{5}{36} (x^{3} - 8)^{\frac{6}{5}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{5}{6} u^{\frac{6}{5}}

= \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = x^{3} - 8

= \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{5}{6} (x^{3} - 8)^{\frac{6}{5}}

Simplify

\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{5}{6} (x^{3} - 8)^{\frac{6}{5}} : \frac{5}{36} (x^{3} - 8)^{\frac{6}{5}}

= \frac{5}{36} (x^{3} - 8)^{\frac{6}{5}}

Add a constant to the solution

= \frac{5}{36} (x^{3} - 8)^{\frac{6}{5}} + C

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What is the integral of 1/2 x^2\sqrt[5]{x^3-8} ?
The integral of 1/2 x^2\sqrt[5]{x^3-8} is 5/36 (x^3-8)^{6/5}+C