Given power series, say:$$f(x)= c_0+c_1(x-a)+c_2(x-a)^2+...= \sum_{n=0}^{\infty}c_n(x-a)^n $$
and when we integrate it we get the following:
$$\int f(x) dx=C+c_0(x-a)+\frac{c_1(x-a)^2}{2}+.... $$
My question is:
Why do we get "$c_0(x-a)$" instead of $c_0x$ given that we've differentiated with respect to x?
On
$\int f(x) dx=\int (c_0+c_1(x-a)+c_2(x-a)^2+...)dx=C'+c_0x+c_1\frac{x^2}{2}-c_1ax+c_2\frac{x^3}{3}-c_2ax^2+c_2a^2x+...=\left(C'-c_0a-c_1\frac{a^2}{2}-c_2\frac{a^3}{3}+...\right) + c_0(x-a)+\frac{c_1(x-a)^2}{2}+\frac{c_2(x-a)^3}{3}+...=C+ c_0(x-a)+\frac{c_1(x-a)^2}{2}+\frac{c_2(x-a)^3}{3}+...$
Where we defined $C=\left(C'-c_0a-c_1\frac{a^2}{2}-c_2\frac{a^3}{3}+...\right)$
You get$$\int f(x)\,\mathrm dx=C+c_0(x-a)+\frac12c_1(x-a)^2+\cdots$$if you are interested in an antiderivative of $f$ which is expressed as a power series centered at $a$, which is natural here, since $f$ is expressed as such a power series. Nevertheless, it is also true that$$\int f(x)\,\mathrm dx=C^\ast+c_0x+\frac12c_1x^2+\cdots$$for some other constant $C^\ast$.