$$\frac{e^x-1}{x(e^{2x}+1)}=\frac{1}{2x\cosh x}-\frac{1}{2x}(1-\tanh x)=\frac{1}{2x}\big(\frac{1}{\cosh x}+\tanh x-1\big)$$
Therefore, integrating by part,
$$I=\int_0^\infty\frac{e^x-1}{x(e^{2x}+1)}dx$$
$$=\frac{\ln x}{2}\big(\frac{1}{\cosh x}+\tanh x-1\big)\bigg|_0^\infty+\frac{1}{2}\int_0^\infty\frac{\ln x\,\sinh x}{\cosh ^2x}dx-\frac{1}{2}\int_0^\infty\frac{\ln x}{\cosh ^2x}dx$$
$$=\frac{1}{2}\int_0^\infty\frac{\ln x\,\sinh x}{\cosh ^2x}dx-\frac{1}{2}\int_0^\infty\frac{\ln x}{\cosh ^2x}dx=I_1+I_2\tag{1}$$
Adding and extracting an additional term and integrating by part the first term
$\displaystyle 2I_1=\int_0^\infty\ln x\Big(\frac{\sinh x}{\cosh^2 x}-e^{-x}\Big)dx+\int_0^\infty\ln x\,e^{-x}dx\tag*{}$
$\displaystyle =-\gamma+\int_0^\infty\Big(\frac{1}{\cosh x}-e^{-x}\Big)\frac{dx}{x}=-\gamma+\int_0^\infty e^{-x}\tanh x\frac{dx}{x}\tag*{}$
Making the substitution $e^{-2x}=t$
$\displaystyle 2I_1=-\gamma+\int_0^1\frac{t^{-\frac{1}{2}}}{\ln t}\,\frac{t-1}{t+1}dt=-\gamma+J\Big(-\frac{1}{2}\Big)\tag*{}$
where we defined
$\displaystyle J(a)=\int_0^1\frac{t^a}{\ln t}\,\frac{t-1}{t+1}dt=\ln(1+a)+2\ln\frac{\Gamma(1+a)}{\Gamma^2(1+a/2)}-(2\ln2)a+\ln\frac{\pi}{2},\,\,a>-1\,\,\tag*{}$
The evaluation of this integral is in the Addendum below. Letting $a=-\frac{1}{2}$
$\displaystyle J\Big(-\frac{1}{2}\Big)=2\ln\pi-\ln2-4\ln\Gamma\Big(\frac{3}{4}\Big)\tag*{}$
and
$\displaystyle I_1=-\frac{\gamma}{2}+\ln\pi-\frac{\ln2}{2}-2\ln\Gamma\Big(\frac{3}{4}\Big)\tag{2}$
Now we are evaluating $I_2$: $\displaystyle I_2=-\frac{1}{2}\int_0^\infty\frac{\ln x}{\cosh^2x}dx\tag*{}=-\frac{\pi}{4}\int_{-\infty}^\infty\frac{\ln(\pi x)}{\cosh^2(\pi x)}dx$
$\displaystyle I_2=-\frac{\ln \pi}{4}\int_{-\infty}^\infty\frac{d x}{\cosh^2x}-\frac{\pi}{4}\int_{-\infty}^\infty\frac{\ln x}{\cosh^2(\pi x)}dx=I_{2A}+I_{2B}\tag{3}$
$$I_{2A}=-\frac{\ln \pi}{4}\int_{-\infty}^\infty\frac{d x}{\cosh^2x}=-\frac{\ln \pi}{4}\tanh x\bigg|_{-\infty}^\infty=-\frac{\ln \pi}{2}\tag{4}$$
To evaluate $I_{2B}$ we note that
$$I_{2B}=-\frac{\pi}{8}\int_{-\infty}^\infty\frac{\ln (x^2)}{\cosh^2(\pi x)}dx=-\frac{\pi}{4}\Re\int_{-\infty}^\infty\frac{\ln (-ix)}{\cosh^2(\pi x)}dx$$
Using the fact that
$$\ln a=\ln\Gamma(x+1)-\ln\Gamma(x)\,\,\text{and that}\,\,\cosh^2(\pi(x+i))=\cosh^2(\pi x)$$
we can present the integral as the integral in the complex plane along the closed rectangular contour: $-R\to R\to R+i\to -R+i\to-R$ (the integrals along the added side paths $R\to R+i$ and $-R+i\to-R$ tend to zero at $R\to\infty$).
$$I_{2B}=-\frac{\pi}{4}\Re\int_{-\infty}^\infty\frac{\ln (-ix)}{\cosh^2(\pi x)}dx=\frac{\pi}{4}\Re\int_{-\infty}^\infty\frac{\ln \Gamma(-ix)}{\cosh^2(\pi x)}dx-\frac{\pi}{4}\Re\int_{-\infty}^\infty\frac{\ln \Gamma(-i(x+i))}{\cosh^2(\pi x)}dx$$
$$=\frac{\pi}{4}\Re\oint\frac{\ln \Gamma(-iz)}{\cosh^2(\pi z)}dz=\frac{\pi}{4}\Re \,\bigg(2\pi i\underset{z=i/2}{\operatorname{Res}}\,\frac{\ln \Gamma(-iz)}{\cosh^2(\pi z)}\bigg)$$
(because we have one double pole inside the contour). The residue evaluation is straightforward:
$$I_{2B}=-\frac{1}{2}\psi\big(\frac{1}{2}\big)=\frac{\gamma}{2}+\ln 2\tag{5}$$
Putting (4) and (5) into (3), and (3) and (2) into (1), we get:
$$I=I_1+I_2=-2\Gamma\Big(\frac{3}{4}\Big)+\frac{\ln 2}{2}+\frac{\ln \pi}{2}\tag{6}$$
Using the Euler' reflection formula $\Gamma\big(\frac{1}{4}\big)\Gamma\big(\frac{3}{4}\big)=\frac{\pi}{\sin\frac{\pi}{4}}=\sqrt2\,\pi$, we get the desired result.
$\bf{Addendum}$
The integrand does not have singularities on the interval; taking the derivative with respect to $a$
$\displaystyle J'(a)=\int_0^1\frac{t^a(t-1)}{(1+t)}dt=\int_0^1 t^adt-2\int_0^1\frac{t^a}{1+t}dt=\frac{1}{1+a}+J_0\tag*{}$
where
$\displaystyle J_0=-2\int_0^1(1-t+t^2-t^3+...)t^a\,dt=-2\Big(\frac{1}{1+a}-\frac{1}{2+a}+\frac{1}{3+a}-...\Big)\tag*{}$
$\displaystyle =-2\lim_{N\to\infty}\Big(\sum_{k=1}^N\frac{1}{k+a}-\sum_{k=1}^{N/2}\frac{1}{k+a/2}\Big)\tag*{}$
Given that $\psi(1+a)=-\gamma-\lim_{N\to\infty}\sum_{k=1}^N \Big(\frac{1}{k+a}-\frac{1}{k}\Big)\,;\,\,\psi(x)$ - digamma-function.
$\displaystyle J(a)=-2\Big(-\psi(1+a)+\psi(1+a/2)+\lim_{N\to\infty}\sum_{k=N/2+1}^N\frac{1}{k}\Big)=2\Big(\psi(1+a)-\psi(1+a/2)-\ln2\Big)\tag*{}$
Using $\psi(x)=\frac{d}{dx}\ln\Gamma(x)$ and taking the antiderivative
$\displaystyle J(a)=\int^aJ'(x)dx+C=\ln(1+a)+2\ln\Gamma(1+a)-4\ln\Gamma(1+a/2)-(2\ln2)a+C\tag*{}$
where $C$ is some constant. To define $C$ we see that at $a\to\infty\,\, I(a)\to0$.
Using the asymptotic $\Gamma(1+a)=\sqrt{2\pi a}\big(\frac{a}{e}\big)^a\,$ at $\,a\to\infty$
$\displaystyle J(a)\sim\ln a-2\ln\sqrt{2\pi}+\ln a-2a\ln a-2a\tag*{}$
$\displaystyle +\ln a+4\ln\sqrt{2\pi}-2\ln (a/2)-2a\ln (a/2)+2a-(2\ln2)a+C\tag*{}$
$\displaystyle J(a\to\infty)\to-2\ln\sqrt{2\pi}+2\ln2+C=\ln\frac{2}{\pi}+\,C=0\tag*{}$
$\displaystyle \,\,\Rightarrow\,\,\boxed{\,\,C=\ln\frac{\pi}{2}\,\,}\tag*{}$
Finally, we get for $J(a)$
$\displaystyle \boxed{\,\,J(a)=\ln(1+a)+2\ln\frac{\Gamma(1+a)}{\Gamma^2(1+a/2)}-(2\ln2)a+\ln\frac{\pi}{2}\,\,}\tag*{}$
