Mathematics

# Evaluate the following integral:$\int { \cfrac { { \left( 1+\sqrt { x } \right) }^{ 2 } }{ \sqrt { x } } } dx$

##### SOLUTION
Let
$t=1+\sqrt{x}\Rightarrow\,dt=\dfrac{1}{2\sqrt{x}}dx$

$\Rightarrow\,2dt=\dfrac{1}{\sqrt{x}}dx$

$\displaystyle\int{\dfrac{{\left(1+\sqrt{x}\right)}^{2}}{\sqrt{x}}dx}$

$=2\displaystyle\int{{t}^{2}\,dt}$

$=\dfrac{2{t}^{3}}{3}+c$

$=\dfrac{2{\left(1+\sqrt{x}\right)}^{3}}{3}+c$ where $t=1+\sqrt{x}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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