TITLE: How to explain why Integration by parts apparently "fails" in the case of $\int \frac{f'(x)}{f(x)}dx$ without resorting to definite integrals?
QUESTION [6 upvotes]: Integrating by parts the following integral $$I=\int \frac{f'(x)}{f(x)}dx$$
gives us
$$\begin{align*}
I&=\int \frac{f'(x)}{f(x)}\,dx\\
&=\int\frac1{f(x)}f'(x)\,dx\\
&=\frac1{f(x)}f(x)-\int\left(\frac1{f(x)}\right)'f(x)\,dx\\
&=1+\int \frac{f'(x)}{f(x)}\,dx,
\end{align*}
$$
so
$$I=1+I\implies0=1,$$
which seems like a contradiction but is in reality a mistake as we can see by being somewhat more rigorous:
$$
\begin{align*}
I&=\int_{a}^x \frac{f'(t)}{f(t)}\,dt\\
&=\int_{a}^x\frac1{f(t)}f'(t)\,dt\\
&=\left[\frac1{f(t)}f(t)\right]_a^x-\int_{a}^x\left(\frac1{f(t)}\right)'f(t)\,dt,
\end{align*}$$
so $I=I.$

How do I explain this to a student-of economics for what it's worth-who has not still learned about definite Integrals? 
(I suspect I could insert some constant on the upper part of the "failed" relations but I am not sure where or how to properly explain it. I also understand that in a couple of lessons we will talk about definitive integrals but what can I say now-That indefinite integrals are in reality a form of definite ones? )

REPLY [4 votes]: It isn't a paradox at all.  Evaluating an indefinite integral amounts to finding an antiderivative of the integrand; i.e., $$\int f(x) \, dx = F(x) \iff F'(x) = f(x).$$  Therefore, the antiderivative is unique only up to addition by a constant.*  Different methods of integration may yield different antiderivatives.
If we think of differentiation as an operator on a set of differentiable functions, then integration is merely an inverse operator, much like taking a square root is the inverse operation of squaring.  But there is no guarantee that such operations are one-to-one:  indeed, if I say $x^2 = 4$, there are two real numbers for which this is true, namely $x = 2$ and $x = -2$.  That this is the case does not imply that we must have $2 = -2$.

*Moreover, one can create antiderivatives that differ by an arbitrary constant on different intervals of the integrand's domain; e.g., the function $$F(x) = \begin{cases} \log x + C, & x > 0 \\ \log(-x) + K, & x < 0, \end{cases}$$ satisfies the relationship $F'(x) = \frac{1}{x}$ for all $x \ne 0$, for any constant choices of $C, K$.  Thus the so-called "constant of integration" does not itself encompass all antiderivatives of a given function; one could have different constants of integration for each interval of an integrand on which it is defined.