Prove
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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