It is important to remember that the u-v substitutions for "Integration by Parts" comes from the following identity for derivatives:
$$
\Big(U(x)\cdot V(x)\Big)' \ = \ U(x)\cdot V'(x) \ + \ U'(x) \cdot V(x)
$$
If you integrate both sides of that expression, and remember that $\int{F'(x) dx}=F(x)$ simplifies the left hand side, you are left with:
$$
U(x)\cdot V(x) \ =\int{U(x)\cdot V'(x) \ dx} \ + \ \int{U'(x)\cdot V(x) \ dx}
$$
that rearranges into:
$$
\int{U(x)\cdot V'(x) \ dx} \ = \ U(x)\cdot V(x)\ - \ \int{U'(x)\cdot V(x) \ dx}
$$
which is the form we usually see the identity in.
When you repeat that process you have to be careful with your new change of variables, so lets call them $Y(x)$ and $Z(x)$ to be clear.
$$
\int{U(x)\cdot V'(x) \ dx} \ = \ U(x)\cdot V(x)\ - \ \int{Y(x)\cdot Z'(x) \ dx}
$$
The "Integration by Parts" identity then gives us:
$$
\int{U(x)\cdot V'(x) \ dx} \ = \ U(x)\cdot V(x)\ - \ \Big( \ Y(x)\cdot Z(x)\ - \ \int{Y'(x)\cdot Z(x) \ dx}\Big)
$$
or rather
$$
\int{U(x)\cdot V'(x) \ dx} \ = \Big( \ U(x)\cdot V(x)\ - \ Y(x)\cdot Z(x)\Big) \ + \ \int{Y'(x)\cdot Z(x) \ dx}
$$
which is all good and correct thusfar, but it doesn't tell us whether we chose $Y(x)=V(x)$ or $Y(x)=U'(x)$ for the substitution. If you choose $Y(x)=V(x)$ then $Z'(x)=U'(x)$ and you can maybe start to see why you're getting $I=I$:
$$
\int{U(x)\cdot V'(x) \ dx} \ = \Big( \ U(x)\cdot V(x)\ - \ V(x)\cdot U(x)\Big) \ + \ \int{V'(x)\cdot U(x) \ dx}
$$
It's like Gerry Myerson said in the comments, that choice of substitution is essentially the reverse of the previous step.
On the other hand, if $Y(x)=U'(x)$ then $Z'(x)=V(x)$ and so you get:
$$
\int{U(x)\cdot V'(x) \ dx} \ = \Big( \ U(x)\cdot V(x)\ - \ U'(x)\cdot \int V(x)\ dx \Big) \ + \ \int{\Big(U''(x)\cdot \int V(x)\ dx \Big)\ dx }
$$
which is actually making forward progress by taking deeper-and-deeper derivatives of $U(x)$ and higher-and-higher anti-derivatives of $V'(x)$.
(Swapping which variable gets the derivative midway through was essentially taking a derivative-then-antiderivative of $U(x)$ and an antiderivative-then-derivative of $V'(x)$... which is exactly why it all cancelled out.)
