What is the integral of (3x^3-3x)^{1/2}(3x^2-1) ?

The integral of (3x^3-3x)^{1/2}(3x^2-1) is 2/9 (3x^3-3x)^{3/2}+C

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integral of (3x^3-3x)^{1/2}(3x^2-1)

\int (3 x^{3} - 3 x)^{\frac{1}{2}} (3 x^{2} - 1) d x

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

Solution steps

\int (3 x^{3} - 3 x)^{\frac{1}{2}} (3 x^{2} - 1) d x

Apply u-substitution

= \int \frac{u ^{\frac{1}{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 u^{\frac{1}{2}} d u

Apply the Power Rule

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

Substitute back

u = 3 x^{3} - 3 x

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

Simplify

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

= \frac{2}{9} (3 x^{3} - 3 x)^{\frac{3}{2}}

Add a constant to the solution

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

What is the integral of (3x^3-3x)^{1/2}(3x^2-1) ?
The integral of (3x^3-3x)^{1/2}(3x^2-1) is 2/9 (3x^3-3x)^{3/2}+C