Evaluar la integral definida

$$\begin{align}&\int_0^2x^3\sqrt{3x+1}dx=\\ &\\ &t=3x+1\\ &dt = 3dx\\ &x= \frac{t-1}{3}\\ &\\ &x=0 \implies t=1\\ &x=2 \implies t=7\\ &\\ &=\frac 13\int_1^7 \frac{(t-1)^3}{27} \sqrt t dt=\\ &\\ &\frac {1}{81}\int_1^7(t^3-3t^2+3t-1)\sqrt t\; dt=\\ &\\ &\frac 1{81}\int(t^{7/2}-3t^{5/2}+3t^{3/2}-t^{1/2})dt=\\ &\\ &\frac{1}{81}\left[\frac 29t^{9/2}-\frac 67 t^{7/2}+\frac 65 t^{5/2}-\frac 23t^{3/2} \right]_1^7=\\ &\\ &\frac{1}{81}\left(\frac 297^4 \sqrt 7-\frac 677^3 \sqrt 7+\frac 65 49 \sqrt 7-\frac 237 \sqrt 7-\frac 29+\frac 67-\frac 65+\frac 23  \right)\\ &\\ &\frac 1{81}\left(\frac{4802 \sqrt 7}{9}-294 \sqrt 7+\frac{294 \sqrt 7}{5} -\frac{14 \sqrt 7}3+\frac{-70+270-378+210}{315}  \right)=\\ &\\ &\frac{1}{81}\left(\frac{(24010-13230+2646-210)\sqrt 7}{45}+\frac{32}{315}\right)=\\ &\\ &\frac{1}{81}\left(\frac{13216 \sqrt 7}{45}+\frac{32}{315}  \right)=\\ &\\ &\frac{13216 \sqrt 7}{3645}+ \frac{32}{25515}\end{align}$$

¡Uff qué cuentas!

Y eso es todo.