Using Integration By Parts results in 0 = 1 I've run into a strange situation while trying to apply Integration By Parts, and I can't seem to come up with an explanation.  I start with the following equation:
$$\int \frac{1}{f} \frac{df}{dx} dx$$
I let:
$$u = \frac{1}{f} \text{ and } dv = \frac{df}{dx} dx$$
Then I find:
$$du = -\frac{1}{f^2} \frac{df}{dx} dx \text{ and } v = f$$
I can then substitute into the usual IBP formula:
$$\int udv = uv - \int v du$$
$$\int \frac{1}{f} \frac{df}{dx} dx = \frac{1}{f} f - \int f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$
$$\int \frac{1}{f} \frac{df}{dx} dx = 1 + \int \frac{1}{f} \frac{df}{dx} dx$$
Then subtracting the integral from both sides, I've now shown that:
$$0 = 1$$
Obviously there must be a problem in my derivation here...  What wrong assumption have I made, or what error have I made?  I'm baffled.
 A: Hint: Constant of integration.
A: You have correctly derived that $0 = 1$... modulo constants.
Antiderivatives are only well-defined modulo constants*; e.g. both $x$ and $x+1$ are antiderivatives (with respect to $x$) of $1$. The equation you wrote is implicitly only meant to be an equation modulo constants; that is, the two sides of the equation don't have to be equal: they're allowed to differ by a constant.
This is traditionally worked around by adding a "constant of integration" in an ad-hoc manner rather than trying to introduce modular arithmetic. This ad-hoc fix can be tricky to get right in a nontrivial algebraic calculation if you don't fully understand what's going on, as your calculation shows.
When you cancel out the two copies of $\int\frac{1}{f} \frac{df}{dx} \, dx$, that doesn't get rid of the fact that the equation is still only meant to hold modulo constants: you've merely eliminated your mental cue (the presence of an antiderivative) to remind you of that fact.
*: Technically, I should say "locally constant functions in the integration variable" rather than "constants"
A: This line
$$\int \frac{1}{f} \frac{df}{dx} dx = \frac{1}{f} f - \int f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$
should be
$$\int_a^b \frac{1}{f} \frac{df}{dx} dx = \left[\frac{1}{f} f\right]_a^b - \int_a^b f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$
so
$$\int_a^b \frac{1}{f} \frac{df}{dx} dx = \left[1\right]_a^b - \int_a^b f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$
and $\left[1\right]_a^b=0$
A: As an exercise, look at this proof of $0=1$
Differentiate both sides $wrt $ $x$
$0=0$
Which is true, hence proved.
If you can find the error here, so you can in your above question. 
A: The problem here is that when you applied the By Parts Formula 
$$uv-\int \frac{du}{dx}vdv   $$
you took $u=\frac 1f$ and $dv=\frac {df}{dx}dx$ Now when you use by parts formula your first task is to get $v$ and for that you need to do this $\int dv$, right? You have done all righty, but the problem comes when you write $\int df =f$
Here you also need to add the constant C, Or your proof will get pretty much messed up. In the end where you got $0=1$ , you are lacking that $C$ in this equation here, when you do antiderivatives you can't just provide one answer.If you write $\int df=f$ only then you are only getting one of the infinite solutions because $\frac {d}{df}(f)=1$ but $\frac {d}{df}(f+2)$ is also 1, infact take the derivative of $f+c$ (w.r.t f) where $C$ is any  constant, you will always end up with 1.So I guess that's the error.Next time be careful with constants.
Hope It answered your question.                                
