Integration of the following What is the definite integral of 
$$
\int_0^1 \left(\frac{g(x)}{f(x)}\right)'\cdot\frac{1}{g(x)}\,dx,
$$
where the conditions are as follows:
$f(0) = 2 $
$f(1) = 3 $
$f'(x) $ is continuous
For all $x$ in the range $\{0,1\}, f(x)^2-g(x)^2=1$.
Thank you very much.
 A: By integration by parts, 
\begin{align}\int_0^1 \left(\frac{g}{f}\right)' \frac{1}{g}\, dx &= \frac{g}{f}\cdot \frac{1}{g}\bigg|_{t = 0}^{t = 1} - \int_0^1 \frac{g}{f}\frac{d}{dx}\left(\frac{1}{g}\right)\, dx\\
&= \frac{1}{f(1)} - \frac{1}{f(0)} + \int_0^1 \frac{g'}{fg}\, dx\\
&= -\frac{1}{6} + \int_0^1 \frac{g'}{fg}\, dx.
\end{align}
Differentiating both sides of the equation $f^2 - g^2 = 1$, we get $2ff' - 2gg' = 0$, or $ff' = gg'$. So $g'/f = f'/g$, in which case $$\int_0^1 \frac{g'}{fg}\, dx = \int_0^1 \int_0^1 \frac{f'}{g^2}\, dx = \int_0^1 \frac{f'}{f^2 - 1}\, dx = \int_{2}^{3} \frac{du}{u^2 - 1} = \frac{1}{2}\log\left|\frac{u - 1}{u + 1}\right|\bigg|_{u = 2}^{u = 3} = \frac{1}{2}\log\frac{3}{2}.$$
Hence $$\int_0^1 \left(\frac{g}{f}\right)'\frac{1}{g}\, dx = -\frac{1}{6} + \frac{1}{2}\log\frac{3}{2}.$$
A: Integrating by parts we get $$\int_0^1\frac{1}{g}d(\frac{g}{f})=\frac{1}{f(1)}-\frac{1}{f(0)}+\int_0^1\frac{gdg}{fg^2}$$
Differentiating $f^2-g^2=1$ and replacing in the integral in the RHS of the above $gdg$ par $fdf$ et $g^2$ par $f^2-1$ we get $$\int_0^1\frac{1}{g}d(\frac{g}{f})=-\frac{1}{6}+\int_0^1\frac{fdf}{f^2-1}$$
