What is the integral of 6x^2sqrt(3x^3-1) ?

The integral of 6x^2sqrt(3x^3-1) is 4/9 (3x^3-1)^{3/2}+C

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integral of 6x^2sqrt(3x^3-1)

\int 6 x^{2} 3 x^{3} - 1 d x

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

Solution steps

\int 6 x^{2} 3 x^{3} - 1 d x

Take the constant out:

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

= 6 \cdot \int x^{2} 3 x^{3} - 1 d x

Apply u-substitution

= 6 \cdot \int \frac{u}{9} d u

Take the constant out:

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

= 6 \cdot \frac{1}{9} \cdot \int u d u

Apply the Power Rule

= 6 \cdot \frac{1}{9} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 x^{3} - 1

= 6 \cdot \frac{1}{9} \cdot \frac{2}{3} (3 x^{3} - 1)^{\frac{3}{2}}

Simplify

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

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

Add a constant to the solution

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

What is the integral of 6x^2sqrt(3x^3-1) ?
The integral of 6x^2sqrt(3x^3-1) is 4/9 (3x^3-1)^{3/2}+C