Question

find its anti derivative ​

Answer 1

Given ,

The function is $\tt \sqrt{x} + \frac{1}{ \sqrt{x} }$

Integrating wrt x , we get

$\tt \implies \int \{\sqrt{x} + \frac{1}{ \sqrt{x} } \} \: dx$

$\tt \implies\int \{ {(x)}^{ \frac{1}{2} } + {(x)}^{ - \frac{1}{2} } \} \: dx$

$\tt \implies\int \{ {(x)}^{ \frac{1}{2} } \} \: \: dx \: + \: \int \{{(x)}^{ - \frac{1}{2} } \}\: dx$

$\tt \implies \frac{ 2{(x)}^{ \frac{3}{2} } }{3} + \frac{2 {(x)}^{ \frac{1}{2} } }{1} + C$

Therefore , the correct option is C

Remmember :

$\tt \implies \int {(x)}^{n} \: \: dx = \frac{ {(x)}^{n + 1} }{n + 1} + C$

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Answer 2

We have,

$\displaystyle\longrightarrow\int x^n\ dx=\dfrac{x^{n+1}}{n+1},\quad\!n\neq-1$

Replacing $n$ by $\dfrac{1}{n}$ we get the following.

$\displaystyle\longrightarrow\int x^{\frac{1}{n}}\ dx=\dfrac{n}{n+1}\cdot x^{\frac{n+1}{n}},\quad\!n\neq-1\quad\quad\dots(i)$

Here,

$\displaystyle\longrightarrow I=\int\left(\sqrt x+\dfrac{1}{\sqrt x}\right)\ dx$

Or,

$\displaystyle\longrightarrow I=\int\left(x^{\frac{1}{2}}+\dfrac{1}{x^{\frac{1}{2}}}\right)\ dx$

$\displaystyle\longrightarrow I=\int\left(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right)\ dx$

$\displaystyle\longrightarrow I=\int x^{\frac{1}{2}}\ dx+\int x^{\frac{1}{-2}}\ dx\quad\quad\dots(1)$

Evaluating each integral using (i),

$\displaystyle\longrightarrow\int x^{\frac{1}{2}}\ dx=\dfrac{2}{3}\,x^{\frac{3}{2}}$

and,

$\displaystyle\longrightarrow\int x^{\frac{1}{-2}}\ dx=\dfrac{-2}{-1}\,x^{\frac{-1}{-2}}$

$\displaystyle\longrightarrow\int x^{\frac{1}{-2}}\ dx=2x^{\frac{1}{2}}$

Then (1) becomes,

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{2}{3}\,x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C}}$

Hence (C) is the answer.