Question:
The anti derivative of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ equals
(A) $\frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+\mathrm{C}$
(B) $\frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^{2}+\mathrm{C}$
(C) $\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+\mathrm{C}$
(D) $\frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+\mathrm{C}$
Solution:
$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x$
$=\int x^{\frac{1}{2}} d x+\int x^{-\frac{1}{2}} d x$
$=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+\mathrm{C}$
$=\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+\mathrm{C}$
Hence, the correct answer is C.
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