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

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

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integral of xsqrt(3x^2-6)

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

\frac{1}{9} (3 x^{2} - 6)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

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

= \int \frac{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 u d u

Apply the Power Rule

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

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

Substitute back

u = 3 x^{2} - 6

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

Simplify

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

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

Add a constant to the solution

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

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\frac{\partial}{\partial x} (x^{4} - 8 x y - 5 y^{2})

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