Help understanding integration $ \int f(ax + b)dx = \frac 1a F(ax + b) + C $

Question

Let F be a antiderivative of the function $f$. Then:

$ \int f(ax + b)dx = \frac 1a F(ax + b) + C $

when a and b are constants and a is not zero

Answer 1

That is the chain rule, in the very simple case of an affine change of variable, from the point of view of antiderivatives.

Answer 2

Let $u = ax+b$. Then $du = a\,dx \Longrightarrow \frac{1}{a}\,du = dx$, and $$\int f(ax + b)dx = \int f(u)\cdot\frac{1}{a}\,du = \frac 1a F(u) + C = \frac 1a F(ax + b) + C.$$