Integrate the functions in Exercises: x x+4 ,x>0

Question

Integrate the functions in Exercises:
$\frac{\text{x}}{\sqrt{\text{x}+4}},\text{x}>0$

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Answer

$ \int\frac{\text{x}}{\sqrt{\text{x + 4}}}\text{ dx}= \int\frac{\text{x}+4-4}{\sqrt{\text{x}+4}}\text{ dx}$
$=\int\frac{\text{x}+4}{\sqrt{\text{x}+ 4}}-\frac{4}{\sqrt{\text{x}+4}}\text{ dx}$
$=\int\sqrt{\text{x}+4}\text{ dx}-4\int\frac{1}{\sqrt{\text{x + 4}}}\text{ dx} $
$=\int(\text{x}+4)^{\frac{1}{2}}\text{ dx}-4\int(\text{x}+ 4)^{\frac{-1}{2}}\text{ dx} $
$=\frac{(\text{x}+4)^{\frac{3}{2}}}{\frac{3}{2}(1)}-\frac{4(\text{x}+4)^{\frac{1}{2}}}{\frac{1}{2}(1)}+\text{c}$
$=\frac{2}{3}(\text{x}+4)^{\frac{3}{2}}-8(\text{x}+ 4)^{\frac{1}{2}}+\text{c}$
$=\frac{2}{3}\sqrt{\text{x}+4}\bigg(\frac{\text{x}+4}{3}-4\bigg)+\text{c} $
$=\frac{2}{3}\sqrt{\text{x}+4}\bigg(\frac{\text{x}+4-12}{3}\bigg)+\text{c} $
$=\frac{2}{3}\sqrt{\text{x}+4}(\text{x}-8)+\text{c} $

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