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?

I am going to evaluate the integral by differentiating its partner
$$
\begin{aligned}
I(a) &=\int_0^1 \frac{x^a-x^{n-1}}{\left(1+x^p\right) \ln x} dx\\
I^{\prime}(a) &=\int_0^1 \frac{x^a}{1+x^p} d x \\
&=\sum_{k=0}^{\infty} \frac{(-1)^k}{a+pk+1}
\end{aligned}
$$
Integrating back gives our integral
$$
\begin{aligned}
I &=I(m-1)-I(n-1)  \\
&=\int_{n-1}^{m-1} \sum_{k=0}^{\infty} \frac{(-1)^k}{a+p k+1} d a \\
&=\sum_{k=0}^{\infty}(-1)^k \ln \left(\frac{m+p k}{n+p k}\right)\\&= \ln \left(\frac{m}{n} \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}…\right)
\end{aligned}
$$
