The *indefinite integral* of a function f(x): J->**R**, where J is some interval, is any continuous function F(x): J->**R**->**R** such that F'(x)=f(x) whenever f is continuous at x. (Such an F(x) is also termed an *antiderivative* of f(x), particularly in older works).

The fundamental theorem of calculus is that the definite integral may be computed in terms of F(x):

∫_{a}^{b} f(x)dx = F(b)-F(a)

(Indeed, this furnishes an alternative

definition of the indefinite integral.)

Note that if F(x) is an indefinite integral of f(x) and `c`∈**R** then F(x)+`c` is also an indefinite integral, as whenever F(x) is differentiable, (F(x)+`c`)'=F'(x)+`c`'=F'(x)+0=f(x). Note also that the formula of the fundamental theorem of calculus continues to hold for F(x)+`c`.

One way to arrive at the indefinite integral of f(x) is to compute the definite integral: the function (of x, defined for some fixed `a`∈J and any x∈J)

F(x)=∫_{a}^{x} f(t)dt

is an indefinite integral of f(x), due to the fundamental theorem. Choosing

`b` rather than

`a` merely gives another indefinite integral F(x)-F(

`b`)+F(

`a`).

(However, not all indefinite integrals F(x) are necessarily of this form; it may be that F(x)+`c` is not of this form, as the magnitude of `c` is too large.)
Computing indefinite integrals is *hard*. Whenever the indefinite integral is "known", *all* definite integrals are "known". And whenever *all* definite integrals are "known", the indefinite integral is "known". But it may (and does!) happen that only *certain* definite integrals are "known" (i.e. may be expressed as an elementary function of constants such as ,1,π and e), and no indefinite integral exists that is an elementary function. These are sometimes (misleadingly) called "impossible integrals".