Normally, integrals look like $\int{f(x)dx}$. But, I occasionally get something like $\int{dx\over 1+f(x)dx}$. Is it a sign of an error? Partcularly, I tried to relativistic constant force acceleration, using speed addition formula
$s(t+dt) = {s(t) + dv \over 1+s(t)\:dv/c^2}$
so that
$\int ds = \int {(s(t+dt) - s(t))} = \int {({s(t) + dv \over 1+s(t)\:dv/c^2} - s(t))}$
and, I cannot factor $dv$ out
edit I have got an answer from Harald Hanche-Olsen, which says that my expression can be reduced to the canonical form s(v)*dv by power series decomposition and discarding all terms $dv^n, n > 1$. Particlularly, $$\int {dx \over 1+f(x)\:dx} = \int {dx (1-f(x)\:dx)} = \int {(dx -f(x)\:dx^2)} = \int dx = x$$
But, I could not understand if you call an integral (and can put under integration sign "$\int$")
- only canonical expression of the form $f(x)dx$ or
- any expression that is reducible to it or
- we can reduce any infinitesimal expression to canonical form by Harald Hanche-Olsen method or
- any infinitecimal expression can be integrated, despite not all of them are reducible to $f(x)dx$?
What you're doing goes counter to how mathematics is usually done, but it does make some sense. If you think of a differential like $dv$ as the limiting case of a small change $\Delta v$, then as $\Delta v\to0$ then higher powers like $\Delta v^2$ become negligible compared to $\Delta v$ itself. Formally, you can write this insight as $dv^2=0$. With this rule in mind, you get $$ s(t+dt) = \frac{s(t) + dv}{1+s(t)\,dv/c^2}=(s(t)+dv)(1-s(t)\,dv/c^2)=s(t)+(1-s(t)^2/c^2)\,dv.$$ What we're really doing here is applying the rule for differentiating a quotient, though.
As for what can be integrated, assuming we are talking about a single variable, it has be on the form $f(x)\,dx$ or at least be reducible to that form.