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? 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? ) 


Integrating by parts the following integral $$I=\int \frac{f'(x)}{f(x)}dx$$
gives us
$$I=\int \frac{f'(x)}{f(x)}dx=\int\frac1{f(x)}f'(x)dx=\\\frac1{f(x)}f(x)-\int\Big(\frac1{f(x)}\Big)'f(x)dx=\\1+\int \frac{f'(x)}{f(x)}dx\Rightarrow\\
I=1+I$$

Be careful here.  The $I$ on the left and the $I$ on the right are not exactly the same quantity.
How can that make sense?  It makes sense because indefinite integrals have an arbitrary constant of integration, and for any given indefinite integral, its arbitrary constant is not guaranteed or required to have the same value everywhere the integral is written.
In fact, the process you followed shows that the arbitrary constant for the $I$ on the left is necessarily $1 + $ the arbitrary constant for the $I$ on the right.
