What is the integral of sqrt(6x) ?

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

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integral of sqrt(6x)

\int 6 x d x

\frac{2 2}{3} x^{\frac{3}{2}} + C

Solution steps

\int 6 x d x

Apply radical rule:

n ab = n a n b,

assuming

a \geq 0, b \geq 0

= \int 6 x d x

Take the constant out:

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

= 6 \cdot \int x d x

Apply the Power Rule

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

= 6 \frac{2}{3} x^{\frac{3}{2}}

Simplify

6 \frac{2}{3} x^{\frac{3}{2}} : \frac{2 2}{3} x^{\frac{3}{2}}

= \frac{2 2}{3} x^{\frac{3}{2}}

Add a constant to the solution

= \frac{2 2}{3} x^{\frac{3}{2}} + C

\int \frac{x + 1}{x ^{2} - 4 x + 3} d x

y = x^{2} - 2, y = 7

f (x) = \frac{- 6 x}{x ^{2} + 1}, (0, 0)

x^{2} + 4

\int \frac{1}{sin ( \frac{y}{2} )} d y

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