Question

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

• A. $\frac{1}{3}x^{\frac{1}{3}}+2x^{\frac{1}{2}}+C$
• B. $\frac{2}{3}x^{\frac{2}{3}}+\frac{1}{2}x^{\frac{1}{2}}+C$
• C. $\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C$
• D. $\frac{2}{3}x^{\frac{3}{2}}+\frac{1}{2}x^{\frac{1}{2}}+C$

Answer 1

Answer: (C)

$\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C$

Answer 2

$\bigg(\sqrt x+\frac{1}{\sqrt x}\bigg)$

= $\int x^{\frac{1}{2}}dx+\int x^{\frac{1}{2}}dx$

= $\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C$

=$\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C$

Hence, the correct Answer is C.