product of integral of reciprocal functions let us consider the integral of the following question :

this question seems quit  interesting for me  and  that why i have decided to think about, this, first of all i  was thinking to use  function like 
$\frac{1}{x}$  
because if $f(x)=\frac{1}{x}$
then definitely $\frac{1}{f(x)}$ is equal to $x$, integral of $\frac{1}{x}=\frac{1}{-x^2}+c$ 
and integral of $x$ is  $\frac{x^2}{2}+c$, but this does not match required condition, should i use this property?

what about  in general  function $f(x)=e^{k*x}$ ?
 A: The notation in this question is rather a red herring, suggesting a sort of spurious symmetry between the antiderivatives of $f$ and $1/f$. Let's remove the ambiguity by writing $F(x)=\int f(x) \, dx$. (Standard rant about antiderivative notation goes here.) Now the equation looks like
$$ F(x)\int \frac{dx}{f(x)} = -1. $$
Dividing through by $F(x)$ (which cannot be identically zero since then $1/f$ makes no sense), we have
$$ \int \frac{dx}{f(x)} = -\frac{1}{F(x)}. $$
Now differentiate:
$$ \frac{1}{f(x)} = \frac{1}{F'(x)} = \frac{F'(x)}{F(x)^2}. $$
Now we can solve this equation for $F$, and hence $f$. We have
$$ 0 = F'^2-F^2 = (F'-F)(F'+F) $$
Hence we have either $F'=F$ or $F'=-F$, so the solution is $F(x)=Ae^x$ or $F(x)=Ae^{-x}$. We must have the same solution for all $x$ since $F(x)$ is never zero, so $F'$ cannot change from $F$ to $-F$ without being undefined at a point. Differentiating gives the final answer
$$ f(x) = Ae^{x} \quad \text{or} \quad f(x) = Ae^{-x}. $$
A: Looking through functions that are easily integrate-able, there is one set of solutions:
$$f(x)=e^{ax}$$
where $|a|=1$, and $a$ can be a complex number.
A: Differentiating both sides we get,$$f\int\dfrac{1}{f}dx + \frac{1}{f}\int{f}dx=0 \implies f^2\int\frac{1}{f}dx+\int fdx=0$$
Differentiate both sides again to get,$$f+ 2ff'\int\frac{1}{f}dx + f=0$$ $$\implies 2ff'\int\frac{1}{f}dx=2f\implies \int\frac{1}{f}dx=\frac{1}{f'}\implies\frac{1}{f}=\frac{f''}{f'^2}$$ $$\implies\frac{f'}{f}=\frac{f''}{f'}\implies lnf=ln(f')+A\implies\frac{f'}{f}=B \implies ln(f)=Bx+C\implies f=e^{Bx+C}$$
