What is the integral of sqrt(6t) ?

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

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integral of sqrt(6t)

\int 6 t d t

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

Solution steps

\int 6 t d t

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \int 6 t d t

Take the constant out:

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

= 6 \cdot \int t d t

Apply the Power Rule

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (5 y - 2 x)

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 100 cos (8 t)

\int x^{2} + x - 6 d x

(2 x - 1) d x + (7 y + 5) d y = 0

\frac{d}{d x} (2 e^{x - 1})

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