Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$

Hint. One may find $a,b \in \mathbb{R}$ such that
$$
 (2x+3) \sqrt {3x+1}=\color{red}{a}\cdot(3x+1)^{3/2}+\color{red}{b}\cdot(3x+1)^{1/2}.
$$
