alleryThe antiderivative of $\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)$ equals
A$\dfrac{1}{3}x^{1/3}+2x^{1/2}+C$
B$\dfrac{2}{3}x^{2/3}+\dfrac{1}{2}x^{2}+C$
C$\dfrac{2}{3}x^{3/2}+2x^{1/2}+C$
D$\dfrac{3}{2}x^{3/2}+\dfrac{1}{2}x^{1/2}+C$
Answer & Solution
Correct answer: C. $\dfrac{2}{3}x^{3/2}+2x^{1/2}+C$
1. Write the integrand as $x^{1/2}+x^{-1/2}$.
2. $\int x^{1/2}\,dx=\dfrac{x^{3/2}}{3/2}=\dfrac{2}{3}x^{3/2}$.
3. $\int x^{-1/2}\,dx=\dfrac{x^{1/2}}{1/2}=2x^{1/2}$.
4. Sum: $\dfrac{2}{3}x^{3/2}+2x^{1/2}+C$. Differentiating returns $x^{1/2}+x^{-1/2}$, confirming C.
_Source: syllabus Mathematics Ch 7 "Integrals", p.296_
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