Calculus - indefinite integration

This is not easier than expanding using the Binomial theorem, but it's a different way to approach it which you may at least find interesting, and even potentially useful (in other situations if not this one).

For any integer $n \ge 0$, let

$$f(n) = \int x(x^3 + 1)^n dx \tag{1}\label{eq1}$$

For $n \ge 1$, using integration by parts, where $u(x) = (x^3 + 1)^n$ so $d(u(x)) = 3nx^2(x^3 + 1)^{n-1}dx$, and $d(v(x)) = xdx$ so $v(x) = \frac{x^2}{2}$, you get

$$\begin{equation}\begin{aligned} f(n) & = \frac{x^2}{2}(x^3 + 1)^n - \frac{3n}{2} \int x^4(x^3 + 1)^{n-1} dx \\ & = \frac{x^2}{2}(x^3 + 1)^n - \frac{3n}{2} \int x(x^3 + 1 - 1)(x^3 + 1)^{n-1} dx \\ & = \frac{x^2}{2}(x^3 + 1)^n - \frac{3n}{2} \int \left(x(x^3 + 1)^n - x(x^3 + 1)^{n-1}\right) dx \\ & = \frac{x^2}{2}(x^3 + 1)^n - \frac{3n}{2} \left(f(n) - f(n-1)\right) \end{aligned}\end{equation}\tag{2}\label{eq2}$$

This leads to the recursive equation

$$\begin{equation}\begin{aligned} \left(1 + \frac{3n}{2}\right)f(n) & = \frac{x^2}{2}(x^3 + 1)^n + \frac{3n}{2}f(n-1) \\ f(n) & = \frac{x^2}{2 + 3n}(x^3 + 1)^n + \frac{3n}{3n + 2}f(n-1) \end{aligned}\end{equation}\tag{3}\label{eq3}$$

You can determine what $f(0)$ is (I'm leaving that to you) and then use \eqref{eq3} to determine each of the rest of the $f$ values up to $f(33)$.