$\displaystyle x=\frac{(2t+3)^\frac{3}{2}}{3}$, $\displaystyle y=t+\frac{t^2}{2}$, $\displaystyle 0\leq t\leq 3$ They want me to find the Length

right now ive gotten to $\displaystyle L=\int^3_0 \sqrt{t^2+5t+4} dt$ I dont know if this is even right, and I dont know how to take this integral either :-(