What is the integral of (x^2)/(\sqrt[3]{2x^3+9)} ?

The integral of (x^2)/(\sqrt[3]{2x^3+9)} is 1/4 (2x^3+9)^{2/3}+C

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integral of (x^2)/(\sqrt[3]{2x^3+9)}

\frac{1}{4} (2 x^{3} + 9)^{\frac{2}{3}} + C

Solution steps

Apply u-substitution

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{3}{2} v^{\frac{2}{3}}

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

Substitute back

= \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{3}{2} (2 x^{3} + 9)^{\frac{2}{3}}

Simplify

\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{3}{2} (2 x^{3} + 9)^{\frac{2}{3}} : \frac{1}{4} (2 x^{3} + 9)^{\frac{2}{3}}

= \frac{1}{4} (2 x^{3} + 9)^{\frac{2}{3}}

Add a constant to the solution

= \frac{1}{4} (2 x^{3} + 9)^{\frac{2}{3}} + C

y^{^{'}} = (3 t + 1)^{2}

f (x) = \frac{2 x}{- x + 2}

\frac{4 d u}{d t} = u^{2}

\int \frac{1}{x ( x ^{2} + 8 ) ^{2}} d x

What is the integral of (x^2)/(\sqrt[3]{2x^3+9)} ?
The integral of (x^2)/(\sqrt[3]{2x^3+9)} is 1/4 (2x^3+9)^{2/3}+C