Question
The anti derivative of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ equals

Answer

$\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x$
= $\int x^{\frac{1}{2}} d x+\int x^{-\frac{1}{2}} d x$
= $\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C$
= $\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C$

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