What is the integral of sqrt(9x-36) ?

The integral of sqrt(9x-36) is 2(x-4)^{3/2}+C

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integral of sqrt(9x-36)

\int 9 x - 36 d x

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

Solution steps

\int 9 x - 36 d x

Apply u-substitution

: \int \frac{u}{9} d u

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 9 x - 36

= \frac{1}{9} \cdot \frac{2}{3} (9 x - 36)^{\frac{3}{2}}

Simplify

\frac{1}{9} \cdot \frac{2}{3} (9 x - 36)^{\frac{3}{2}} : 2 (x - 4)^{\frac{3}{2}}

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

Add a constant to the solution

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

\frac{d y}{d x} = 10 e^{x - y}

x^{^{''}} = x

x^{2} - 1 arcsec (x)

\frac{\partial}{\partial y} (2 e^{y} x)

x^{2} y^{^{'}} + 2 x y + 1 = 0

What is the integral of sqrt(9x-36) ?
The integral of sqrt(9x-36) is 2(x-4)^{3/2}+C