What is the relationship between $\int f(x) \, dx$ and $\int 1/f(x) \, dx$? So many integration problems would be made so easy if we were just allowed to work with the reciprocal of a function. Is there a way to relate $\int f(x) \, dx$ and $\int 1/f(x) \, dx$ ? How about if you had limits of integration? Thanks.
P.S. If there is no relationship, is it proven that there is no relationship or has someone just not found it yet? 
 A: If $F(x) = \int f(x)\,dx$ and $G(x) = \int dx/f(x)$, then the relationship between them is:
$$
F'(x) G'(x) = 1 .
$$
If this is of no use to you (as I suspect), then perhaps you need to specify what "no relationship" means.
A: Note that for $x\gt0$, the minimum of $x+\frac1x$ is $2$; that is,
$$
x+\frac1x-2=\left(\sqrt{x}-\frac1{\sqrt{x}}\right)^2\ge0
$$
Therefore, we get for $f\gt0$,
$$
\int_a^b\frac1{f(x)}\mathrm{d}x+\int_a^bf(x)\,\mathrm{d}x\ge2(b-a)
$$
A: Since $x\mapsto 1/x$ is convex when $x\gt0$, we have by Jensen's Inequality
$$
\frac1{b-a}\int_a^b\frac1{f(x)}\mathrm{d}x\ge\frac1{\frac1{b-a}\int_a^bf(x)\,\mathrm{d}x}
$$
when $f\gt0$, which implies
$$
\int_a^b\frac1{f(x)}\mathrm{d}x\int_a^bf(x)\,\mathrm{d}x\ge(b-a)^2
$$
Looking at the preceding inequality, it also follows directly from Hölder's Inequality.
A: Here is a precise sense in which no relation is to be expected: $\displaystyle \int\log(x)\,dx$ can be easily computed, and found to be $x\log(x)-x+C$. On the other hand, $\displaystyle \int\frac{dx}{\log(x)}$ is non-elementary, that is, it admits no closed form expression in terms of elementary functions. (A good reference for this is Bronstein's book on Symbolic integration.) This is to say, in general, you will not find a "useful" expression for $\int F(x)\,dx$ in terms of $\int dx/F(x)$, or vice versa.
Similarly, we can give examples where one definite integral converges while the other diverges, so there is no "useful" relation in the sense of definite integrals either. (We can easy show inequalities involving them, though, for example using Cauchy-Schwarz, or as in robjohn's answer.)
