I have played around with the differences between definite and indefinite integrals and noted the following:
$$\int C_{0}+C_{1}x\,{\rm d}x-\int_{0}^{x}C_{0}+C_{1}t\,{\rm d}t=0$$
$$\int\ln(x)\,{\rm d}x-\int_{0}^{x}\ln(t)\,{\rm d}t=0$$
$$\int\exp(x)\,{\rm d}x-\int_{0}^{x}\exp(t)\,{\rm d}t=1$$
$$\int x\,\exp(x)\,{\rm d}x-\int_{0}^{x}t\,\exp(t)\,{\rm d}t=-1$$
$$\int\sinh(C\, x)\,{\rm d}x-\int_{0}^{x}\sinh(C\, t)\,{\rm d}t=1/C$$
$$\int\sinh^{-1}(x)\,{\rm d}x-\int_{0}^{x}\sinh^{-1}(t)\,{\rm d}t=-1$$
$$\int\cosh^{-1}(x)\,{\rm d}x-\int_{0}^{x}\cosh^{-1}(t)\,{\rm d}t=\mbox{-}{\mathtt{i}}$$
$$\int\sinh^{-1}(C/x)\,{\rm d}x-\int_{0}^{x}\sinh^{-1}(C/t)\,{\rm d}t=C\,\ln(C)$$
and I was wondering if there is anything special with functions whose result is $0$ or not zero. Actually I started from the last one in a numerical method I was developing, and was intrigued by this extra $C\,\ln(C)$ term that I needed when I calculated the integral with definite limits.
For a polynomial it seems the answer is always zero (I have no proof of it), but for other functions it is not. This is interesting since I thought any continuous function can be approximated as a polynomial using a Taylor's series expansion.
So my question is which functions have their indefinite integral equals the definite one, and which not. Does this have to do with so called transcendental functions.
** EDIT: **
I am using a CAS
(computer algebra system) to do the integrations. So maybe the results are implementation specific.