Mathematics

# Integrate: $\displaystyle \int \dfrac{x}{\sqrt{x+4}}dx$

$\dfrac 23(x+4)^{\tfrac 32}-8\sqrt{x+4}$

##### SOLUTION
Given, $I=\displaystyle \int \dfrac{x}{\sqrt{x+4}}dx$

let $u=\sqrt{x+4}$ $\Rightarrow du=\dfrac{1}{\sqrt{x+4}}$

$\Rightarrow \displaystyle \int \:2\left(u^2-4\right)du$

$\displaystyle =2\left(\int \:u^2du-\int \:4du\right)$

$=2\left(\dfrac{u^3}{3}-4u\right)$

$=2\left(\dfrac{\left(\sqrt{x+4}\right)^3}{3}-4\sqrt{x+4}\right)$

$=2\left(\dfrac{1}{3}\left(x+4\right)^{\frac{3}{2}}-4\sqrt{x+4}\right)$

$=\dfrac{2}{3}\left(x+4\right)^{\frac{3}{2}}-8\sqrt{x+4}$

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