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

$\begin{array}{1 1} (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\end{array} $

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1 Answer

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  • $(i)\;\int x^n=\frac{x^n+1}{n+1}+c.$
  • $(ii)\;\frac{d}{dx}[f(x)+c]=f(x)$.
  • $\int f(x)dx=f(x)+c$.
We know that
$\sqrt x=\frac{d}{dx}\bigg(\frac{2}{3}x^\frac{3}{2}\bigg)$
Thus the anti derivative of $\sqrt x$ is $\frac{2}{3}x^\frac{3}{2}$.------(1)
We know that
$\frac{1}{\sqrt x}=\frac{d}{dx}(2x^\frac{1}{2})$.
Thus the anti derivative of $\frac{1}{\sqrt x}=\frac{d}{dx}(2x^\frac{1}{2})$-------(2)
Combining (1) and (2) we get,
$\sqrt x+\frac{1}{\sqrt x}=\frac{2}{3}x^\frac{3}{2}+2x^\frac{1}{2}+c.$
Thus the correct answer is C.


answered Jan 28, 2013 by sreemathi.v
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