What is the integral of x^2*\sqrt[4]{2x^3+9} ?

The integral of x^2*\sqrt[4]{2x^3+9} is 2/15 (2x^3+9)^{5/4}+C

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integral of x^2*\sqrt[4]{2x^3+9}

\int x^{2} \cdot 4 2 x^{3} + 9 d x

\frac{2}{15} (2 x^{3} + 9)^{\frac{5}{4}} + C

Solution steps

\int x^{2} 4 2 x^{3} + 9 d x

Apply u-substitution

: \int \frac{4 u}{6} d u

= \int \frac{4 u}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int 4 u d u

Apply the Power Rule

: \frac{4}{5} u^{\frac{5}{4}}

= \frac{1}{6} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 2 x^{3} + 9

= \frac{1}{6} \cdot \frac{4}{5} (2 x^{3} + 9)^{\frac{5}{4}}

Simplify

\frac{1}{6} \cdot \frac{4}{5} (2 x^{3} + 9)^{\frac{5}{4}} : \frac{2}{15} (2 x^{3} + 9)^{\frac{5}{4}}

= \frac{2}{15} (2 x^{3} + 9)^{\frac{5}{4}}

Add a constant to the solution

= \frac{2}{15} (2 x^{3} + 9)^{\frac{5}{4}} + C

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What is the integral of x^2*\sqrt[4]{2x^3+9} ?
The integral of x^2*\sqrt[4]{2x^3+9} is 2/15 (2x^3+9)^{5/4}+C