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Let $I=\int \sqrt{x^{2}+3 x} d x$

$=\int \sqrt{x^{2}+3 x+\frac{9}{4}-\frac{9}{4}} d x$

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

It is known that, $\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+\mathrm{C}$

$\begin{aligned} \therefore I &=\frac{\left(x+\frac{3}{2}\right)}{2} \sqrt{x^{2}+3 x}-\frac{9}{2} \log \left|\left(x+\frac{3}{2}\right)+\sqrt{x^{2}+3 x}\right|+C \\ &=\frac{(2 x+3)}{4} \sqrt{x^{2}+3 x}-\frac{9}{8} \log \left(x+\frac{3}{2}\right)+\sqrt{x^{2}+3 x}+C \end{aligned}$