The anti derivative of \(\bigg(\sqrt x+\frac{1}{\sqrt x}\bigg)\) equals
\(\frac{1}{3}x^{\frac{1}{3}}+2x^{\frac{1}{2}}+C\)
\(\frac{2}{3}x^{\frac{2}{3}}+\frac{1}{2}x^{\frac{1}{2}}+C\)
\(\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C\)
\(\frac{2}{3}x^{\frac{3}{2}}+\frac{1}{2}x^{\frac{1}{2}}+C\)
\(\bigg(\sqrt x+\frac{1}{\sqrt x}\bigg)\)
= \(\int x^{\frac{1}{2}}dx+\int x^{\frac{1}{2}}dx\)
= \(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C\)
=\(\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C\)
Hence, the correct Answer is C.
What is the Planning Process?
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Integral calculus helps to resolve two major types of problems: