The antiderivative of left(sqrt{x}+frac{1}{sq | vedclass

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Question

(B) We need to find the integral: $\int \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) dx$Rewrite the expression using exponents: $\int \left(x^{\frac{1}{2}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We need to find the integral: $\int \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) dx$Rewrite the expression using exponents: $\int \left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right) dx$Apply the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:For the first term: $\int x^{\frac{1}{2}} dx = \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3} x^{\frac{3}{2}}$For the second term: $\int x^{-\frac{1}{2}} dx = \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} = \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2 x^{\frac{1}{2}}$Combining these results,we get: $\frac{2}{3} x^{\frac{3}{2}} + 2 x^{\frac{1}{2}} + C$,where $C$ is an arbitrary constant.Hence,the correct answer is $B$.

The antiderivative of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ equals

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