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

Let us just split $2x+3 = \frac23 (3x+1)+\frac73$ and simplify to give us: $$I = \int (2x+3)\sqrt{3x+1}\, dx = \frac23 \int (3x+1)^{\frac32}\, dx+ \frac73 \int \sqrt {3x+1}\, dx$$ which can be easily solved.
