What is the integral of 1/(sqrt(2gh)) ?

The integral of 1/(sqrt(2gh)) is (sqrt(2)sqrt(h))/(sqrt(g))+C

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integral of 1/(sqrt(2gh))

\int \frac{1}{2 g h} d h

\frac{2 h}{g} + C

Solution steps

\int \frac{1}{2 g h} d h

2 g h = 2 g h

= \int \frac{1}{2 g h} d h

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{g h} d h

\frac{1}{g h} = \frac{1}{g} \frac{1}{h}

= \frac{1}{2} \cdot \int \frac{1}{g} \cdot \frac{1}{h} d h

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{g} \cdot \int \frac{1}{h} d h

Apply the Power Rule

: 2 h^{\frac{1}{2}}

= \frac{1}{2} \cdot \frac{1}{g} \cdot 2 h^{\frac{1}{2}}

Simplify

\frac{1}{2} \cdot \frac{1}{g} \cdot 2 h^{\frac{1}{2}} : \frac{2 h}{g}

= \frac{2 h}{g}

Add a constant to the solution

= \frac{2 h}{g} + C

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What is the integral of 1/(sqrt(2gh)) ?
The integral of 1/(sqrt(2gh)) is (sqrt(2)sqrt(h))/(sqrt(g))+C