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

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

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integral of sqrt(3x-6)

\int 3 x - 6 d x

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

Solution steps

\int 3 x - 6 d x

Apply u-substitution

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

= \int \frac{u}{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 d u

Apply the Power Rule

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

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

Substitute back

u = 3 x - 6

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

Simplify

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

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

Add a constant to the solution

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

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\frac{\partial}{\partial x} (x^{3} + 2 (5 - 5 x^{2})^{2} + 2)

\frac{\partial}{\partial x} (e^{3 x y})

\int \frac{ln ( 10 x )}{x} d x

\frac{d}{d x} (I n (\frac{4 + x ^{2}}{x}))

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