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The antiderivative of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ equals
(a) $\frac{1}{3} x^{1 / 3}+2 x^{1 / 2}+C$
(b) $\frac{2}{3} x^{2 / 3}+\frac{1}{2} x^2+C$
(c) $\frac{2}{3} x^{3 / 2}+2 x^{1 / 2}+C$
(d) $\frac{3}{2} x^{3 / 2}+\frac{1}{2} x^{1 / 2}+C$
The antiderivative of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ equals
(a) $\frac{1}{3} x^{1 / 3}+2 x^{1 / 2}+C$
(b) $\frac{2}{3} x^{2 / 3}+\frac{1}{2} x^2+C$
(c) $\frac{2}{3} x^{3 / 2}+2 x^{1 / 2}+C$
(d) $\frac{3}{2} x^{3 / 2}+\frac{1}{2} x^{1 / 2}+C$
Solution:
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(c) $\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x=\int\left(x^{1 / 2}+x^{-\frac{1}{2}}\right) d x$
$=\frac{2}{3} x^{3 / 2}+2 x^{1 / 2}+\mathrm{C}$
$=\frac{2}{3} x^{3 / 2}+2 x^{1 / 2}+\mathrm{C}$
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