<s>[INST] 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?
 [/INST] Let $a\geq 0$ and $$f(a)=\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p)\ln x}x^a dx$$
You're really looking for the value of $f(0)$.
Note that
$$\begin{split}
f^\prime(a) &= \int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p)}x^a dx\\
&= \int_0^1 (x^{m-1}-x^{n-1})\left( \sum_{k\geq 0} (-1)^k x^{kp}\right)x^a dx\\
&= \sum_{k\geq 0} (-1)^k\int_0^1 (x^{kp+a+m-1} -x^{kp+a+n-1}) dx\\
&= \sum_{k\geq 0} (-1)^k\left( \frac 1 {kp+a+m}-\frac 1 {kp+a+n}\right)
\end{split}$$
Integrating back w.r.t. $a$, and noting that $\lim_{a\rightarrow+\infty}f(a)=0$ gives
$$\begin{split}
f(a) &= \sum_{k\geq 0} (-1)^k \ln\left( \frac{kp + a + m}{kp + a + n}\right)\\
&=\ln\left ( \frac{a+m}{a+n} \cdot \frac{p+a+n}{p+a+m}\cdot \frac{2p+a+m}{2p+a+n}\dots\right)
\end{split}
$$
Evaluating at $a=0$:
$$\int_0^1\frac{x^{m-1}-x^{n-1}}{(1+x^p)\ln x} dx= \ln\left ( \frac{m}{n} \cdot \frac{p+n}{p+m}\cdot \frac{2p+m}{2p+n}\dots\right)$$
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