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

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

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integral of 3sqrt(x+9)

\int 3 x + 9 d x

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

Solution steps

\int 3 x + 9 d x

Take the constant out:

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

= 3 \cdot \int x + 9 d x

Apply u-substitution

= 3 \cdot \int u d u

Apply the Power Rule

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

Substitute back

u = x + 9

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

Simplify

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

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

Add a constant to the solution

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

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