Solve ∫ dh/sqrt{2gsqrt{h}}

$$\int \frac { d h } { \sqrt { 2 g } \sqrt { h } }$$

Evaluate

$$\sqrt{\frac{2h}{g}}+С$$

Solution Steps

Factor out the constant using \int af\left(h\right)\mathrm{d}h=a\int f\left(h\right)\mathrm{d}h.

$$\frac{\int \frac{1}{\sqrt{h}}\mathrm{d}h}{\sqrt{2}\sqrt{g}}$$

Rewrite \frac{1}{\sqrt{h}} as h^{-\frac{1}{2}}. Since \int h^{k}\mathrm{d}h=\frac{h^{k+1}}{k+1} for k\neq -1, replace \int h^{-\frac{1}{2}}\mathrm{d}h with \frac{h^{\frac{1}{2}}}{\frac{1}{2}}. Simplify and convert from exponential to radical form.

$$\frac{2\sqrt{h}}{\sqrt{2}\sqrt{g}}$$

Simplify.

$$\frac{\sqrt{h}\sqrt{2}}{\sqrt{g}}$$

If F\left(h\right) is an antiderivative of f\left(h\right), then the set of all antiderivatives of f\left(h\right) is given by F\left(h\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{\sqrt{h}\sqrt{2}}{\sqrt{g}}+С$$

Differentiate w.r.t. g

$$-\frac{\sqrt{2h}}{2g^{\frac{3}{2}}}$$