<s>[INST] Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$
 [/INST] By parts: $u=2x+3$ and $\mathrm dv=\sqrt{3x+1}$, then 
$$
\int(2x+3)\sqrt{3x+1}\;\mathrm dx=\frac{2}{9}(2x+3)(3x+1)^{3/2}-\frac{4}{9}\int(3x+1)^{3/2}\;\mathrm dx\\
=\frac{2}{9}(2x+3)(3x+1)^{3/2}-\frac{8}{135}(3x+1)^{5/2}+C
$$
</s>