Question

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

• A. $\frac{2}{3}(x+4{)}^{\frac{3}{2}}-8\sqrt{x+4}$
• B. $\frac{2}{3}(x+4{)}^{\frac{3}{2}}+8\sqrt{x+4}$
• C. $\frac{2}{3}(x+4{)}^{\frac{3}{2}}+4\sqrt{x+4}$
• D. None of these

Answer 1

The correct option is A $\frac{2}{3}(x+4{)}^{\frac{3}{2}}-8\sqrt{x+4}$

Given, $I=\int \frac{x}{\sqrt{x+4}}dx$

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

$\Rightarrow \int 2(u^2-4)du$

$=2(\int u^{2}du-\int 4du)$

$=2(\frac{u^3}{3}-4u)$

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

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

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