Question

find antiderivative of:
$\sqrt{x}$
+1/
$\sqrt{x}$
​

Answer 1

Solution:-

$\rm \: \to \int \bigg( \sqrt{x} + \dfrac{1}{ \sqrt{x} } \bigg)dx$

$\rm \: \to \int \bigg( {x}^{ \frac{1}{2} } + \dfrac{1}{ {x}^{ \frac{1}{2} } } \bigg)dx$

Using this properties

$\rm \implies \: \int \{f(x) + g(x) \}dx = \int f(x)dx + \int g(x)dx$

We get

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

$\rm \: \to \: \dfrac{ {x}^{ \frac{1}{2} + 1 } }{ \dfrac{1}{2} + 1 } + \dfrac{ {x}^{ \frac{ - 1}{2} + 1 } }{ - \dfrac{ 1}{2} + 1 } + c$

$\rm \: \to \: \dfrac{x {}^{ \frac{1 + 2}{2} } }{ \dfrac{1 + 2}{2} } + \dfrac{x {}^{ \frac{ - 1 + 2}{2} } }{ \dfrac{ - 1 + 2}{2} } + c$

$\to \rm \: \dfrac{x {}^{ \frac{3}{2} } }{ \dfrac{3}{2} } + \dfrac{x {}^{ \frac{1}{2} } }{ \dfrac{1}{2} } + c$

$\rm \to \: \dfrac{2x {}^{ \frac{3}{2} } }{3} + \dfrac{2x {}^{ \frac{1}{2} } }{1} + c$

Answer

$\rm \to \: \dfrac{2x {}^{ \frac{3}{2} } }{3} + \dfrac{2x {}^{ \frac{1}{2} } }{1} + c$