So I noticed that there's an occasion when integration by substitution is "simpler" than integration by substitution.
The integral I'm having is
$$\int_{x_0}^x (t-x_0)^n dt, \space t, x_0, x \in \mathbb{R}, n \in \mathbb{N}$$
Now to use integration by substitution on this one defines $f(t)=t^n$ and $g(t)=t-x_0$ and writes
$$=\int_{x_0}^x (t-x_0)^n \cdot 1\space dt$$
$$= \int_{x_0}^{x} f(g(t))g'(t)dt$$
And by integration by substitution formula
$$=\int_{g(x_0)}^{g(x)} f(u) du$$
$$=\int_{0}^{x-x_0} f(u) du$$
$$=\int_{0}^{x-x_0} u^n du$$
And it turns out that in this case one can use the power formula
$$\int u^n = \frac{u^{n+1}}{n+1} + C_1, \space u\in \mathbb{R}, n \in \mathbb{N}$$
Now the question is, is this some general property of certain kinds of integral of powers? That one can directly use the power rule interpreted like
$$\int f(x)^n dx = \frac{f(x)^{n+1}}{n+1} + C_2$$
or
$$\int x^n dx = \frac{x^{n+1}}{n+1} + C_3, \space x \in \mathbb{R}$$
rather than use integration by substitution?
Or does the inner function have to be of first order (with or without added constants and without a multiplicative constant acting on the variable) so that one gets the inner function derivative $g'(x)=1$?