Mathematics

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


ANSWER

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


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
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