Show $\int_\frac{1}{3}^\frac{1}{2}\frac{\operatorname{artanh}(t)}{t}dt=\int_{\ln 2}^{\ln 3}\frac{u}{2\sinh u}du$ How would I show (or explain) that
$$\int_\frac{1}{3}^\frac{1}{2}\frac{\operatorname{artanh} t}{t}dt,$$
$$\int_{\ln 2}^{\ln 3}\frac{u}{2\sinh u}du,$$
and
$$-\int_\frac{1}{3}^\frac{1}{2}\frac{\ln v}{1-v^2}dv$$
are all equivalent, without evaluating any of the integrals?
 A: As Théophile suggested in the comments, the limits of integration provide a natural starting point: $\ln\frac12=-\ln 2$, $\ln\frac13=-\ln 3$, and $$-\int_\frac{1}{3}^\frac{1}{2}\frac{\ln v}{1-v^2}dv=\int_\frac{1}{2}^\frac{1}{3}\frac{\ln v}{1-v^2}dv\;,$$ so one should consider the possibility that $u=-\ln v$. If so, $du=-\frac1v dv$, and $$\sinh u=\frac12\left(e^u-e^{-u}\right)=\frac12\left(e^{-\ln v}-e^{\ln v}\right)=\frac12\left(\frac1v-v\right)=\frac{1-v^2}{2v}\;,$$ so
$$\int_{\ln 2}^{\ln 3}\frac{u}{2\sinh u}du=\int_{\frac12}^{\frac13}\frac{-\ln v}{\frac{1-v^2}v}\left(-\frac1v\right)dv=\int_{\frac13}^{\frac12}\frac{\ln v}{1-v^2}dv=-\int_{\frac12}^{\frac13}\frac{\ln v}{1-v^2}dv\;.$$
The other equalities will also succumb to reasonable substitutions.
A: I'd like to give a suggestion for the equality
$\displaystyle \int_{1/3}^{1/2} \frac{\text{arctanh } t}{t}\text{ d}t=-\int_{1/3}^{1/2} \frac{\log v}{1-v^2}\text{ d}v$
The idea is to rewrite $\displaystyle \text{arctanh } t=\int_0^t \frac{1}{1-s^2}\text{ d}s$, so that we have
$\displaystyle \int_{1/3}^{1/2} \frac{\text{arctanh } t}{t}\text{ d}t=\int_{1/3}^{1/2} \int_0^t \frac{1}{1-s^2} \frac{1}{t} \text{ d}s \text{ d}t$
Now we can change the order of integration (note that the region isn't "simple"):
$\displaystyle \int_{1/3}^{1/2} \int_0^t \frac{1}{1-s^2} \frac{1}{t} \text{ d}s \text{ d}t=\int_{0}^{1/3} \int_{1/3}^{1/2} \frac{1}{1-s^2} \frac{1}{t} \text{ d}t \text{ d}s+\int_{1/3}^{1/2} \int_{s}^{1/2} \frac{1}{1-s^2} \frac{1}{t} \text{ d}t \text{ d}s$
It is very easy to explicitly evaluate several of the terms now:
$\begin{align*}
&=\displaystyle \frac{1}{2} \log(3/2) \log(2)+\int_{1/3}^{1/2} \frac{\log(1/2)-\log s}{1-s^2}\text{ d}s \\
&=\frac{1}{2} \log(3/2) \log(2)+\int_{1/3}^{1/2} \frac{\log(1/2)}{1-s^2}\text{ d}s - \int_{1/3}^{1/2} \frac{\log s}{1-s^2}\text{ d}s \\
&=\frac{1}{2} \log(3/2) \log(2)-\frac{1}{2} \log(3/2) \log(2) - \int_{1/3}^{1/2} \frac{\log s}{1-s^2}\text{ d}s \\
&=- \int_{1/3}^{1/2} \frac{\log s}{1-s^2}\text{ d}s
\end{align*}$
This is precisely the desired term!
