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.

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.                                
