Is there a function $f(x,a)$ such that $\int_0^1 1/f(x,a) dx = \ln a$ Is there a $f$ such that
\begin{align} 
\int_0^1 \frac{1}{f(x,a)} dx = \ln a
\end{align}
Where the numerator is any constant.
The best I could do:
\begin{align} 
\int_0^1 \frac{1}{1+x(a-1)} dx = \frac{\ln a}{a-1}
\end{align}
Some clarification... the numerator inside the integral must be a constant that does not relay neither on $x$ or $a$, so multiplying the above by $a-1$ is not allowed.
 A: The answer is yes :
$$f(x,a)=\frac{1}{\ln(a)}$$
works unless $a = 1$.
Also you can modify your example to:
$$\frac{a-1}{1+x(a-1)}$$
A: We can start with the assumption that the function has the form $f(x,a) = {x + g(a)}$ and see if we can find $g(a)$. Of course, you can start with a different assumption, but we are only looking for one possible solution.
$\int_0^1 \frac{dx}{f(x,a)} = \int_0^1 \frac{dx}{x+g(a)} = ln \left( 1 + \frac{1}{g(a)} \right)$.
Setting $ln \left( 1 + \frac{1}{g(a)} \right) = ln(a)$, we get $1 + \frac{1}{g(a)} = a$, implying $g(a) = \frac{1}{a-1}$. 
Given our starting assumption, we have $f(x,a) = x + \frac{1}{a-1}$.
A: By Frullani's theorem $f(a,x)=\frac{e^{-x}-e^{-ax}}{x}$ does the job, if the integration range is $\mathbb{R}^+$.
This is the integral representation for $\log$ that leads to the usual integral representation for $\gamma$, for instance.
A: If you have found that
$$\int_0^1 \frac{dx}{1+x(a-1)}=\frac{\ln(a)}{a-1}$$
then just multiply both sides by $(a-1)$ to get
$$\int_0^1 \frac{a-1}{1+x(a-1)}dx=\ln(a)$$
so that we have
$$f(x,a)=\frac{1+x(a-1)}{a-1}=x+\frac{1}{a-1}$$
