Evaluate
2\sqrt{z}+С
Differentiate w.r.t. z
\frac{1}{\sqrt{z}}
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2\sqrt{z}
Rewrite \frac{1}{\sqrt{z}} as z^{-\frac{1}{2}}. Since \int z^{k}\mathrm{d}z=\frac{z^{k+1}}{k+1} for k\neq -1, replace \int z^{-\frac{1}{2}}\mathrm{d}z with \frac{z^{\frac{1}{2}}}{\frac{1}{2}}. Simplify and convert from exponential to radical form.
2\sqrt{z}+С
If F\left(z\right) is an antiderivative of f\left(z\right), then the set of all antiderivatives of f\left(z\right) is given by F\left(z\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.
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