In general, when doing integration by parts, shouldn't you get the same result no matter what "$u$" or "$\mathrm{d}v$" you pick?
I understand the strategy for picking "$u$" and "$\mathrm{d}v$" ("$u$" should be easily differentiable, "$\mathrm{d}v$" should have a ready antiderivative) but there are some cases where the choice is arbitrary: both options for "$u$" and "$\mathrm{d}v$" end up being just fine.
Can anyone help me?
This answer is something of a frame challenge, but I think that you are approaching this (by which I mean, calculus) from the wrong point of view. The question seems to be built on the assumption that the integration by parts formula is merely an algorithm which allows one to transform an integral into another integral for the purposes of finding a useful closed-form antiderivative. This, in turn, seems to be built on the assumption that the goal of calculus is to compute integrals (and derivatives, and series, and whatever). Generally speaking, this is not the right way to view things—if your goal is computation, then $999$ out of $1000$, you are going to be relying on numerical approximations from a computer, anyway, so being able to compute by hand is not really helpful. In reality, the integration by parts formula (and other theorems) are useful for understanding deeper structures and phenomena.
With respect to integration by parts, the reality is that the formula is a statement about the equality of two quantities. Specifically, $$ \int u(x) v'(x) \,\mathrm{d} x = u(x)v(x) - \int u'(x) v(x) \,\mathrm{d}x, $$ which is really just a restatement of the product rule for derivatives, as seen through the lens of the fundamental theorem of calculus. While this can be used to evaluate certain integrals in terms of "nice" closed forms (if, for example, $u'v$ has an "obvious" antiderivative), it is good for so much more.
By way of example, suppose that $f$ is any "nice" function (say, $f$ is smooth in the sense that the $k$-th derivative $f^{(k)}$ exists and is continuous for any $k$). It follows from the fundamental theorem of calculus that for any fixed $x$ and $a$, $$ f(x) - f(a) = \int_{a}^{x} f(t)\,\mathrm{d}t \implies f(x) = f(a) + \int_{a}^{x} f'(t)\,\mathrm{d}t. $$ The integral on the right can be expanded via a clever integration by parts: take $u(t) = f'(t)$ and $v'(t) = 1$. An antiderivative of $v'$ is $t$, but any function of the form $t+C$ will be another antiderivative. So instead of taking $C = 0$ (which is the "usual" approach), the more interesting choice of $C = -x$ gives $v(t) = t-x$.
Then \begin{align} f(x) &= f(a) + \Bigl( f'(t)(t-x) \Bigr]_{t=a}^{x} - \int_{a}^{x} (t-x)f''(t)\,\mathrm{d} t \\ &= f(a) - f'(a)(a-x) - \int_{a}^{x} (t-x)f''(t)\,\mathrm{d}t \\ &= f(a) + (x-a) f'(a) + \int_{a}^{x} (x-t)f''(t)\,\mathrm{d}t. \end{align} This looks more complicated (and, to be fair, maybe it is), but it is possible to keep going in this fashion. Another integration by parts step, this time with $u(t) = f''(t)$ and $v'(t) = x-t$. With this choice of $v'$, $$ v(t) = -\frac{1}{2} t^2 + tx + C = -\frac{1}{2} (t^2 - 2tx + C) = -\frac{1}{2}(t-x)^2, $$ by the choice $C = x^2$ in the third expression. But then $$ f(x) = f(a) + (x-a)f'(a) + \frac{1}{2} (x-a)^2 f''(a) + \frac{1}{2} \int_{a}^{b} (t-x)^2 f'''(t) \,\mathrm{d}t.$$ Repeating this process over and over again eventuall leads to the statement \begin{align} f(x) &= f(a) + (x-a) f'(a) + \dotsb + \frac{1}{k!} (x-a)^k f^{(k)}(a) + \frac{1}{k!} \int_{a}^{x} (t-x)^{k+1} f^{(k+1)}(t)\,\mathrm{d}t \\ &= \sum_{n=0}^{k} \frac{1}{n!} (x-a)^n f^{(n)}(a) + \frac{1}{k!} \int_{a}^{x} (t-x)^{k+1} f^{(k+1)}(t)\,\mathrm{d}t, \end{align} which is a version of Taylor's theorem with an integral remainder term.
If that last integral happens to be small—more precisely, if $$ \lim_{k\to\infty} \left( \frac{1}{k!} \int_{a}^{x} (t-x)^{k+1} f^{(k+1)}(t)\,\mathrm{d}t \right) = 0, $$ then the original function $f$ can be represented by an infinite series. Or, perhaps more practically, the original function $f$ can be well-approximated by a polynomial. In the style of Littlewood's Principles of Analysis, every "nice" function is almost a polynomial.
This is an incredibly powerful observation which follows from an abuse of the integration by parts formula.
If I were always thinking about finding a "nice" $u$ and $\mathrm{d}v$ so that my integration by parts worked out to something cleanly, I would never see this remarkable result. In the argument above, I intentionally chose the "wrong" $u$ and $\mathrm{d}v$ over and over again, yet somehow ended up with something spectacular.
As such, I think that it is worthwhile to change the way that you think about the integration by parts formula (and, in turn, all of calculus). Instead of asking "What choice of $u$ and $\mathrm{d}v$ do I need to make?", you should be asking "What choices of $u$ and $\mathrm{d}v$ give me interesting or useful results?"
To an extent, I think that you are already getting there—the question "don't I get the same thing either way?" indicates that you are starting to think about playing around with these things. However, instead of asking us if there is a "right" answer, I would recommend that you just play around, and see what you get. Explore the ideas that you are learning about, and see what kinds of interesting things pop up!
When picking $u$ and $dv$ for integration by parts, picking either one results in a valid resulting equation, but not necessarily one that is easier to integrate (which is the point). Sometimes either one will work, but usually not.
Here's an example: $\int x\,e^x dx$. If we get it wrong and set $u = e^x$ and $dv = x\,dx$, then $v = \frac{1}{2}x^2$ and $du = e^x\,dx$. Therefore, the equation becomes:
$$ \int x\,e^x\,dx = \frac{1}{2}x^2e^x - \int \frac{1}{2}x^2 e^x dx$$
This is a true statement, but it isn't any more helpful in solving the integral than the original. However, if you set $u = x$ and $dv = e^x\,dx$, that means that $du = dx$ and $v = e^x$, and then this becomes
$$ \int x\,e^x\,dx = x\,e^x - \int e^x\,dx $$
which is extremely easy to solve.
So, either way gives you a "true" result, but one way gets you closer to a solution and the other gets you further away.
For choosing $u$ and $dv$, here's the rules I generally teach.
First of all, I separate out things that "differentiate to zero" (i.e., repeated differentials will eventually remove them, such as polynomial terms) and things that "differentiate cyclically" (i.e., sine, cosine, exponentials, which eventually get back to where they were).
My usual advice is:
There is also a mnemonic that many people use, LIATE, which stands for "logarithmic", "inverse trig", "algebraic", "trig", and "exponential". $u$ gets set to whichever is first on this list, and $dv$ gets the other part.
It is not the case that you will get "the same result".
The point is that 'by parts' relates your original integral $I$ say, to a second integral. If you choose $u$ and $v$ differently you will relate $I$ to a different integral.
The art of using 'by parts' is to make a choice that will result in a new integral that you can actually integrate.
When integrating by parts, you should get the same result no matter what "$u$" or "$\mathrm{d}v$" you choose. For example, the two choices below in integrating $\int \text{sech}^2 x \>\text{csch}^2 x\>dx$ by parts are equally preferred
\begin{align} &\int \text{sech}^2 x \>\text{csch}^2 x\>dx=\int \text{csch}^2 x\>d(\tanh x) =\>\text{csch}^2x\>\tanh x -2\coth x+C\\ &\int \text{sech}^2 x \>\text{csch}^2 x\>dx=-\int \text{sech}^2 x\>d(\coth x) =-\>\text{sech}^2x\>\coth x -2\tanh x+C \end{align}
and the two resulting anti-derivatives, though appearing different, are equivalent.
