What is the integral of sqrt(3x-9) ?

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

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integral of sqrt(3x-9)

\int 3 x - 9 d x

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

Solution steps

\int 3 x - 9 d x

Apply u-substitution

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

Substitute back

u = 3 x - 9

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

Simplify

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

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

Add a constant to the solution

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

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