Integrate the function x^ 2 +3 x - Vidyadip

Question

Integrate the function $\sqrt{x^{2}+3 x}$

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Answer

$I=\int \sqrt{x^{2}+3 x} d x$
= $\int \sqrt{x^{2}+3 x+\frac{9}{4}-\frac{9}{4}} d x$
= $\int \sqrt{\left(x+\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}} d x$
We know that
$\Rightarrow$ $\int \sqrt{x^{2}-a^{2} x} d x$ = $\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$
Therefore,
$I=\frac{\left(x+\frac{3}{2}\right)}{2} \sqrt{x^{2}+3 x}-\frac{9}{8} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}+3 x}\right|+C$
$=\frac{(2 \mathrm{x}+3)}{4} \sqrt{\mathrm{x}^{2}+3 \mathrm{x}}-\frac{9}{8} \log \left|\left(\mathrm{x}+\frac{3}{2}\right)+\sqrt{\mathrm{x}^{2}+3 \mathrm{x}}\right|+\mathrm{C}$

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