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

The integral of x^2sqrt(3x^3-2) is 2/27 (3x^3-2)^{3/2}+C

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

\int x^{2} 3 x^{3} - 2 d x

\frac{2}{27} (3 x^{3} - 2)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 3 x^{3} - 2 d x

Apply u-substitution

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

= \int \frac{3 u - 2}{3} d u

Take the constant out:

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

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

Apply u-substitution

: \int \frac{v}{3} d v

= \frac{1}{3} \cdot \int \frac{v}{3} d v

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{3} \cdot \int v d v

Apply the Power Rule

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

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

Substitute back

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

Simplify

\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} (3 x^{3} - 2)^{\frac{3}{2}} : \frac{2}{27} (3 x^{3} - 2)^{\frac{3}{2}}

= \frac{2}{27} (3 x^{3} - 2)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{27} (3 x^{3} - 2)^{\frac{3}{2}} + C

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What is the integral of x^2sqrt(3x^3-2) ?
The integral of x^2sqrt(3x^3-2) is 2/27 (3x^3-2)^{3/2}+C