Integral of (1+3t)t^2dt dx

1. Rewrite the integrand:

   $$t^{2} \left(3 t + 1\right) = 3 t^{3} + t^{2}$$

2. Integrate term-by-term:

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int 3 t^{3}\, dt = 3 \int t^{3}\, dt$$

      1. The integral of $t^{n}$ is $\frac{t^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int t^{3}\, dt = \frac{t^{4}}{4}$$

      So, the result is: $\frac{3 t^{4}}{4}$

   1. The integral of $t^{n}$ is $\frac{t^{n + 1}}{n + 1}$ when $n \neq -1$:

      $$\int t^{2}\, dt = \frac{t^{3}}{3}$$

   The result is: $\frac{3 t^{4}}{4} + \frac{t^{3}}{3}$

3. Now simplify:

   $$\frac{t^{3} \left(9 t + 4\right)}{12}$$

4. Add the constant of integration:

   $$\frac{t^{3} \left(9 t + 4\right)}{12}+ \mathrm{constant}$$

The answer is:

$$\frac{t^{3} \left(9 t + 4\right)}{12}+ \mathrm{constant}$$