$\int \sqrt{x^2+3 x} \mathrm{~d} x=$

Evaluate the integral: $\int \sqrt{x^2 + 3x} d x$

Complete the square: $x^2 + 3x = \left(x + \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2$.

Let $u = x + \frac{3}{2}$ and $a = \frac{3}{2}$, so the integral becomes $\int \sqrt{u^2 - a^2} d u$.

Using the standard formula $\int \sqrt{u^2 - a^2} d u = \frac{u}{2} \sqrt{u^2 - a^2} - \frac{a^2}{2} \log|u + \sqrt{u^2 - a^2}| + C$:

$I = \frac{x + \frac{3}{2}}{2} \sqrt{x^2 + 3x} - \frac{ \left(\frac{3}{2}\right)^2 }{2} \log\left| x + \frac{3}{2} + \sqrt{x^2 + 3x} \right| + C$

Simplifying:

$I = \frac{2x + 3}{4} \sqrt{x^2 + 3x} - \frac{9}{8} \log\left| x + \frac{3}{2} + \sqrt{x^2 + 3x} \right| + C$

This matches option B.