Prove that $\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln\big(\frac mn \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\dots\big)$ page 371 in ‘Synopsis Of Elementary Results In Pure Mathematics’ contains the following result,
$$\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln \left(\frac{m}{n} \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}…\right)$$
How does one prove this step-by-step?

\begin{align}
&\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln x}\,dx \\
=&\int_0^1 \frac{x^{m -1}-x^{n -1}}{\ln x}\sum_{k\ge 0} (-x^p)^k \,dx\\
=&\ \sum_{k\ge 0}(-1)^k
\int_0^1 \frac{x^{m +pk-1}-x^{n +pk -1}}{\ln x}
\> \ \overset{u=-\ln x}{dx}\\
 =&\ \sum_{k\ge 0} (-1)^k
\int_0^\infty \frac{e^{-(n +pk)u}-e^{-(m +pk)u}}{u}du\\
 =&\ \sum_{k\ge 0} (-1)^k\ln \frac{m+kp}{n+kp}
= \ln \frac{m}{n} \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\cdots
\end{align}
