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

Let $I = \int \sqrt{x^{2}+3x} \, dx$.
Complete the square inside the square root:
$x^{2}+3x = x^{2}+3x + \left(\frac{3}{2}\right)^{2} - \left(\frac{3}{2}\right)^{2} = \left(x+\frac{3}{2}\right)^{2} - \left(\frac{3}{2}\right)^{2}$.
Thus,$I = \int \sqrt{\left(x+\frac{3}{2}\right)^{2} - \left(\frac{3}{2}\right)^{2}} \, dx$.
Using the standard integral formula $\int \sqrt{t^{2}-a^{2}} \, dt = \frac{t}{2} \sqrt{t^{2}-a^{2}} - \frac{a^{2}}{2} \ln |t + \sqrt{t^{2}-a^{2}}| + C$,where $t = x+\frac{3}{2}$ and $a = \frac{3}{2}$:
$I = \frac{x+\frac{3}{2}}{2} \sqrt{\left(x+\frac{3}{2}\right)^{2} - \left(\frac{3}{2}\right)^{2}} - \frac{(\frac{3}{2})^{2}}{2} \ln |(x+\frac{3}{2}) + \sqrt{\left(x+\frac{3}{2}\right)^{2} - \left(\frac{3}{2}\right)^{2}}| + C$.
Simplifying the expression:
$I = \frac{2x+3}{4} \sqrt{x^{2}+3x} - \frac{9}{8} \ln |x+\frac{3}{2} + \sqrt{x^{2}+3x}| + C$.