Question

The anti-derivative of$(\sqrt{x}+\frac{1}{\sqrt{x}})$ equals to (where $C$ is the constant of integration)

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

Answer 1

The correct option is C $\left(\frac{2}{3}\right)x^{\frac{3}{2}}+\left(2\right)x^{\frac{1}{2}}+C$

We know, Antiderivative of $g\left(x\right)$ is $=\int g\left(x\right)dx$
So, antiderivative of given expression is
$=\int (\sqrt{x}+\frac{1}{\sqrt{x}})dx=\int x^{\frac{1}{2}}dx+\int x^{-\frac{1}{2}}dx=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C$
$=\frac{2}{3}{x}^{\frac{3}{2}}+2x^{\frac{1}{2}}+C$