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 $f(x)=e^{k*x}$ ?

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,$$f+ 2ff'\int\frac{1}{f}dx + f=0$$ $$\implies 2ff'\int\frac{1}{f}dx=2f\implies \int\frac{1}{f}dx=\frac{1}{f'}\implies\frac{1}{f}=\frac{f''}{f'^2}$$ $$\implies\frac{f'}{f}=\frac{f''}{f'}\implies lnf=ln(f')+A\implies\frac{f'}{f}=B \implies ln(f)=Bx+C\implies f=e^{Bx+C}$$
