Integral of t^3(1+t^4)^3dt dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = t^{4} + 1$.

      Then let $du = 4 t^{3} dt$ and substitute $\frac{du}{4}$:

      $$\int \frac{u^{3}}{16}\, du$$

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

         $$\int \frac{u^{3}}{4}\, du = \frac{\int u^{3}\, du}{4}$$

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

            $$\int u^{3}\, du = \frac{u^{4}}{4}$$

         So, the result is: $\frac{u^{4}}{16}$

      Now substitute $u$ back in:

      $$\frac{\left(t^{4} + 1\right)^{4}}{16}$$

   Method #2

   1. Rewrite the integrand:

      $$t^{3} \left(t^{4} + 1\right)^{3} \cdot 1 = t^{15} + 3 t^{11} + 3 t^{7} + t^{3}$$

   2. Integrate term-by-term:

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

         $$\int t^{15}\, dt = \frac{t^{16}}{16}$$

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

         $$\int 3 t^{11}\, dt = 3 \int t^{11}\, dt$$

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

            $$\int t^{11}\, dt = \frac{t^{12}}{12}$$

         So, the result is: $\frac{t^{12}}{4}$

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

         $$\int 3 t^{7}\, dt = 3 \int t^{7}\, dt$$

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

            $$\int t^{7}\, dt = \frac{t^{8}}{8}$$

         So, the result is: $\frac{3 t^{8}}{8}$

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

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

      The result is: $\frac{t^{16}}{16} + \frac{t^{12}}{4} + \frac{3 t^{8}}{8} + \frac{t^{4}}{4}$

2. Add the constant of integration:

   $$\frac{\left(t^{4} + 1\right)^{4}}{16}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(t^{4} + 1\right)^{4}}{16}+ \mathrm{constant}$$