Question
Integrate the function: $\frac{x}{\sqrt{x+4}}, x>0$

Answer

Let x + 4 = t
$\Rightarrow$ dx = dt
$\Rightarrow \int \frac{x}{\sqrt{x+4}} d x=\int \frac{(t-4)}{\sqrt{t}} d t$
= $\int\left(\sqrt{t}-\frac{4}{\sqrt{t}}\right) d t$
= $\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+C$
= $\frac{2}{3}(t)^{\frac{3}{2}}-8(t)^{\frac{1}{2}}+C$
= $ \frac{2}{3} t \cdot t^{\frac{1}{2}}-8(t)^{\frac{1}{2}}+C$
= $\frac{2}{3}(x+4)^{\frac{1}{2}}(x+4-12)+C$
= $ \frac{2}{3} \sqrt{x+4}(x-8)+C$.

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