Question

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

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

Answer 1

Answer: (C)

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

$\int (\sqrt{x}+\frac{1}{\sqrt{x}})dx$

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

we know that $\int x^{n}dx=\frac{x^{n+1}}{n+1}$

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

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