Integral of x(x^2+1)^3 dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = x^{2} + 1$.

      Then let $du = 2 x dx$ and substitute $\frac{du}{2}$:

      $$\int \frac{u^{3}}{2}\, du$$

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

         $$\int u^{3}\, du = \frac{\int u^{3}\, du}{2}$$

         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: $\frac{u^{4}}{8}$

      Now substitute $u$ back in:

      $$\frac{\left(x^{2} + 1\right)^{4}}{8}$$

   Method #2

   1. Rewrite the integrand:

      $$x \left(x^{2} + 1\right)^{3} = x^{7} + 3 x^{5} + 3 x^{3} + x$$

   2. Integrate term-by-term:

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

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

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

         $$\int 3 x^{5}\, dx = 3 \int x^{5}\, dx$$

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

            $$\int x^{5}\, dx = \frac{x^{6}}{6}$$

         So, the result is: $\frac{x^{6}}{2}$

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

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

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

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

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

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

         $$\int x\, dx = \frac{x^{2}}{2}$$

      The result is: $\frac{x^{8}}{8} + \frac{x^{6}}{2} + \frac{3 x^{4}}{4} + \frac{x^{2}}{2}$

2. Now simplify:

   $$\frac{\left(x^{2} + 1\right)^{4}}{8}$$

3. Add the constant of integration:

   $$\frac{\left(x^{2} + 1\right)^{4}}{8}+ \mathrm{constant}$$

The answer is: