Calculus - indefinite integration The integral in which I am interested in is
$$\int x(x^3+1)^{33}\mathrm{d}x$$
I tried to solve by substituting $x^2 = t$, but it didn't help. I find a solution by expanding it with the help of binomial expansion. Can anyone help me with any other method like substitution, by parts?

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 $f(33)$.
