Q.Evaluate $\int\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)dx$.

Rewriting each term with a fractional exponent and applying the power-rule integral gives the antiderivative.

$\int x^{1/2}dx=\frac23x^{3/2}$; $\int x^{-1/2}dx=2x^{1/2}$.

Rewriting with exponents

Applying the power rule to each term

Check (differentiate the answer back): $\frac{d}{dx}\left[\frac23x^{3/2}\right]=x^{1/2}=\sqrt{x}$; $\frac{d}{dx}\left[2x^{1/2}\right]=x^{-1/2}=\frac{1}{\sqrt{x}}$; sum $=\sqrt{x}+\frac{1}{\sqrt{x}}$ — exactly the original integrand.

Misapplying the power rule to a negative exponent, e.g. writing $\int x^{-1/2}dx$ as $\frac{x^{-1/2+1}}{-1/2}$ with the wrong sign in the denominator instead of the correct $+1/2$.