Integrating $\int_{0}^{1} x^a (c-x)^b dx $

Question

Problem

How can I integrate $$\int_{0}^{1} x^a (c-x)^b dx $$ where $a,b,c$ are some constants.

Solution Attempt

For $c=1$, I have $\int_{0}^{1} x^a (1-x)^b dx = B(a+1,b+1)$, however, when $c \neq 1$ I am lost.

Answer 1

Enforcing $x =ct$ gives $$\int_{0}^{1} x^a (c-x)^b dx= c^{a+b+1}\int_{0}^{c} t^a (1-t)^b dt =\color{red}{c^{a+b+1} \mathrm {B} (c;\,a,b) }$$

where $\mathrm {B} (c;\,a,b)$is the incomplete beta function, a generalization of the beta function.