Integral of x^3(x^4+1)^5 dx

1. There are multiple ways to do this integral.

   Method #1

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

      Then let $du = 4 x^{3} dx$ and substitute $\frac{du}{4}$:

      $$\int \frac{u^{5}}{4}\, du$$

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

         $$\int u^{5}\, du = \frac{\int u^{5}\, du}{4}$$

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

            $$\int u^{5}\, du = \frac{u^{6}}{6}$$

         So, the result is: $\frac{u^{6}}{24}$

      Now substitute $u$ back in:

      $$\frac{\left(x^{4} + 1\right)^{6}}{24}$$

   Method #2

   1. Rewrite the integrand:

      $$x^{3} \left(x^{4} + 1\right)^{5} = x^{23} + 5 x^{19} + 10 x^{15} + 10 x^{11} + 5 x^{7} + x^{3}$$

   2. Integrate term-by-term:

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

         $$\int x^{23}\, dx = \frac{x^{24}}{24}$$

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

         $$\int 5 x^{19}\, dx = 5 \int x^{19}\, dx$$

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

            $$\int x^{19}\, dx = \frac{x^{20}}{20}$$

         So, the result is: $\frac{x^{20}}{4}$

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

         $$\int 10 x^{15}\, dx = 10 \int x^{15}\, dx$$

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

            $$\int x^{15}\, dx = \frac{x^{16}}{16}$$

         So, the result is: $\frac{5 x^{16}}{8}$

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

         $$\int 10 x^{11}\, dx = 10 \int x^{11}\, dx$$

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

            $$\int x^{11}\, dx = \frac{x^{12}}{12}$$

         So, the result is: $\frac{5 x^{12}}{6}$

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

         $$\int 5 x^{7}\, dx = 5 \int x^{7}\, dx$$

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

            $$\int x^{7}\, dx = \frac{x^{8}}{8}$$

         So, the result is: $\frac{5 x^{8}}{8}$

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

         $$\int x^{3}\, dx = \frac{x^{4}}{4}$$

      The result is: $\frac{x^{24}}{24} + \frac{x^{20}}{4} + \frac{5 x^{16}}{8} + \frac{5 x^{12}}{6} + \frac{5 x^{8}}{8} + \frac{x^{4}}{4}$

2. Now simplify:

   $$\frac{\left(x^{4} + 1\right)^{6}}{24}$$

3. Add the constant of integration:

   $$\frac{\left(x^{4} + 1\right)^{6}}{24}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(x^{4} + 1\right)^{6}}{24}+ \mathrm{constant}$$