Mathematics

# $\displaystyle \int \dfrac {dx}{\sqrt {x}(x+9)}$ is equal to, $(x > 0)$

$\dfrac {2}{3}\tan^{-1}\left(\dfrac {\sqrt {x}}{3}\right)+c$

##### SOLUTION
$\displaystyle \int \dfrac {dx}{\sqrt {x}(x+9)}\ \Rightarrow \sqrt {x}=U$
$\dfrac {1}{2\sqrt {x}}dx=dv\Rightarrow 2vdv=dx$
$\displaystyle \int \dfrac {dx}{U(U^2+9)}-\dfrac {2}{3}\tan^{-1}\left (\dfrac {U}{3}\right)=\dfrac {2}{3}\tan^{-1}\left (\dfrac {\sqrt {x}}{3}\right)$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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