Question
Integrate the function in Exercise.
$\sqrt{\text{x}^2+3\text{x}}$

Answer

$\int\sqrt{\text{x}^2+3\text{x}}\text{dx}=\int\sqrt{\text{x}^2+3\text{x}+\Bigg(\frac{3}{2}\Bigg)^2-\Bigg(\frac{3}{2}\Bigg)^2}\text{dx}=\int\sqrt{\Bigg(\text{x}+\frac{3}{2}\Bigg)^2-\Bigg(\frac{3}{2}\Bigg)^2}\text{dx}$
$=\frac{\text{x}+\frac{3}{2}}{2}\sqrt{\Bigg(\text{x}+\frac{3}{2}\Bigg)^2-\Bigg(\frac{3}{2}\Bigg)^2}-\frac{\Bigg(\frac{3}{2}\Bigg)^2}{2}\text{log}\Bigg|\text{x}+\frac{3}{2}+\sqrt{\Bigg(\text{x}+\frac{3}{2}\Bigg)^2-\Bigg(\frac{3}{2}\Bigg)^2}\Bigg|+\text{c}$
$\Bigg[\therefore\int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2-\text{a}^2}-\frac{\text{a}^2}{2}\text{log}\Bigg|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Bigg|\Bigg]$
$=\frac{2\text{x}+3}{4}\sqrt{\text{x}^2+3\text{x}}-\frac{9}{8}\text{log}\Bigg|\text{x}+\frac{3}{2}+\sqrt{\text{x}^2+3\text{x}}\Bigg|+\text{c}$

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