If one attempts a u-substitution that leaves $x$ in the new integrand (i.e., by reintroducing an $x$ when substituting $dx$ out), it stands to reason that the $x$ could be eliminated by inverting the substitution function to make a further substitution.
For example, to solve $$\int \sqrt{1 + x^{-2}}\ dx$$ one might let $u = 1 + x^{-2}$, and thus $dx = -\frac{x^3}{2}\ du$. The u-substitution would then yield $$\int -\sqrt{u}\ \frac{x^3}{2}\ du.$$
We could solve for u in terms of x by inverting the substitution function to get $x =±\sqrt{\frac{1}{u - 1}}$, and eliminate the final reference to $x$ in the integrand. However, I have never had any luck getting a nice integral at the end of this process. Furthermore, it is generally advised to use u-substitution only when you can see a function and its derivative natively in the integrand. Is there some deeper reason this double substitution approach is doomed to fail (due to the linear dependence of the two substitutions or something), or is worth trying?
Certainly! This is actually the more useful direction, since if you already see the derivative in the integrand, then it's sort of obvious that it's going to work, whereas in this case it's not.
For example, take $$ \int \frac{1}{1+e^x} \, dx , $$ and let $u = 1+e^x$. Then $du = e^x \, dx$, where you didn't have any $e^x$ together with $dx$ in the integral to begin with, so you write the relation as $du = (u-1) dx$ instead and get $$ \int \frac{1}{u} \frac{du}{u-1} , $$ which can be computed using partial fractions.
(In this case, you could also use the little trick of writing the integral as $$ \int \frac{e^{-x}}{e^{-x}+1} \, dx = \int \frac{-dt}{t} $$ where you do have the derivative of $t=e^{-x}+1$ together with $dx$, but the point is that you don't need to notice this in order to succeed, you can simply go ahead and substitute as above.)