Mathematics

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


ANSWER

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


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
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