$\int f(x) dx\cdot \int \frac{1}{f(x)}dx = c$ where $c$ is a constant. Find $f(x)$ 

$$\int f(x) dx\cdot \int \frac{1}{f(x)}dx = c$$ where $c$ is a constant. Find $f(x)$. 


Off first glance it seems that $f(x)$ is some form of $e^x$, but how does one go about doing this analytically (only 1 mark for guessing). I took the derivative of this but the resulting function was still ugly. I thought maybe treating it as a first order linear differential equation would work, but I wasn't sure how to put it in the form $y' + P(x)y = Q(x)$
 A: Let $F(x)$ be the antiderivative of $f(x)$.
$$\int\frac1{F'(x)}dx=\frac c{F(x)},$$
$$\frac1{F'(x)}=-\frac{cF'(x)}{F^2(x)}.$$
Then
$$\left(F'(x)+\frac1{\sqrt{-c}}F(x)\right)\left(\,F'(x)-\frac1{\sqrt{-c}}F(x)\right)=0,$$
has the desired form (and of course the solutions are exponential). There are no real solutions for $c\ge0$.
A: It is given that $$\int\frac{1}{f(x)}dx\cdot \int f(x)dx =c ...(1)$$
Now differentiating both side w.r.t $x$, We Get
$$\int \frac{1}{f(x)}dx\cdot f(x)+\frac{1}{f(x)}\cdot \int f(x)dx =0... (2)$$
Now from equation $(1)$, We get $$ \int\frac{1}{f(x)}dx = \frac{c}{\int f(x)dx}$$
and put into equation $(2)$, We get $$\frac{c\cdot f(x)}{\int f(x)dx}+\frac{\int f(x)dx}{f(x)} = 0$$
So $$ \left(\int f(x)dx\right)^2 = -c\cdot \left(f(x)\right)^2$$
Now Let $-c=\alpha^2\;,$ Then $$\left(\int f(x)dx\right)^2 = \alpha^2\cdot \left(f(x)\right)^2$$
So $$\int f(x)dx = \pm \alpha\cdot f(x)\;,$$ Differentiate both side w.r.t $x\;,$ we get
$$\Rightarrow f(x) = \pm \alpha\cdot f'(x)\Rightarrow \frac{f'(x)}{f(x)}=\pm k\;,$$ where $k = \pm \frac{1}{\alpha}$.
Now Integrate both side w.r.t $x\;,$ we get
$$\Rightarrow \int \frac{f'(x)}{f(x)} = \pm k\int dx\Rightarrow \ln\left|f(x)\right| = \pm k+\ln \left|c\right|\Rightarrow f(x)=c\cdot e^{\pm kx}$$ Hope it helps.
