Fleshing out Ragib's idea: we have the substitution $\,t\to -u+\pi i\Longrightarrow dt=-du,$ , so we put
$$f(t):=\frac{t^2e^t}{1+e^{2t}}\Longrightarrow$$
$$\Longrightarrow\int_{R+\pi i}^{-R+\pi i}f(t)\,dt=-\int_{-R}^R\frac{(u^2-2\pi iu-\pi^2)e^{-u+\pi i}}{1+e^{-2u+2\pi i}}\,du=$$
$$=\int_{-R}^R\frac{u^2e^{-u}}{1+e^{-2u}}du-2\pi i\int_{-R}^R\frac{ue^{-u}}{1+e^{-2u}}\,du-\pi^2\int_{-R}^R\frac{e^{-u}}{1+e^{-2u}}\,du$$
The first integral above is just like the one on the real axis (BTW, note the signs in front of the two last integrals, which come from the factor $\,e^{\pi i}=-1\,$).
The last integral is
$$\pi^2\int_{-R}^R\frac{d(e^{-u})}{1+e^{-2u}}=\left.\pi^2\arctan e^{-u}\right|_{-R}^R=$$
$$=\pi^2\left(\arctan e^{-R}-\arctan e^R\right)\xrightarrow [R\to\infty]{}\pi^2\left(0-\frac{\pi}{2}\right)=-\frac{\pi^3}{2}$$
The second integral above is zero as the integrand is just $\,\frac{\pi iu}{\cosh u}\,$ , which is an odd function...
Denoting our integral by $\, I \,$, we thus have that
$$2I-\frac{\pi ^3}{2}=-\frac{\pi^3}{4}\Longrightarrow I=\frac{\pi^3}{8}$$
$$\int_0^{\infty} \dfrac{\log(x)}{1-x^2} dx = \int_0^{1} \dfrac{\log(x)}{1-x^2} dx + \int_1^{\infty} \dfrac{\log(x)}{1-x^2} dx$$
$$\int_1^{\infty} \dfrac{\log(x)}{1-x^2} dx = \int_1^0 \dfrac{\log(1/x)}{1-1/x^2} \left(-\dfrac{dx}{x^2} \right) = \int_1^0 \dfrac{\log(x)}{x^2-1} dx=\int_0^1 \dfrac{\log(x)}{1-x^2}dx$$
Hence,
$$\int_0^{\infty} \dfrac{\log(x)}{1-x^2} dx = 2\int_0^{1} \dfrac{\log(x)}{1-x^2} dx$$
Now note that in $(0,1)$, we have $$\dfrac1{1-x^2}= \sum_{k=0}^{\infty} x^{2k} \,\,\,\,\,\,\,\, \text{(Geometric/Taylor series)}$$
$$\int_0^{1} \dfrac{\log(x)}{1-x^2} dx = \int_0^1 \left( \sum_{k=0}^{\infty} x^{2k} \right)\log(x) dx = \sum_{k=0}^{\infty} \int_0^1 x^{2k} \log(x) dx = -\sum_{k=0}^{\infty}\dfrac1{(2k+1)^2}$$
(If $\displaystyle \int_0^{1} \dfrac{\log(x)}{1-x^2} dx$, exists, we can afford to swap the integral and summation thanks, to dominated convergence theorem.)
Hence,
$$\int_0^{\infty} \dfrac{\log(x)}{1-x^2} dx = - \sum_{k=0}^{\infty} \dfrac2{(2k+1)^2} =-\dfrac{\pi^2}4$$
Your integral is
$$\int_0^{\infty} \dfrac{\log(x)}{x^2-1} dx = - \int_0^{\infty} \dfrac{\log(x)}{1-x^2} dx = \dfrac{\pi^2}4$$
Best Answer
Take these lines as a long comment and not an answer, but the integral is trivial to compute through real-analytic techniques. For any $a>0$ we have:
$$ \int_{0}^{+\infty}\frac{\log(x/a)}{a^2+x^2}\,dx = \frac{1}{a}\int_{0}^{+\infty}\frac{\log x}{1+x^2}=\color{red}{0} $$ since the substitution $x\mapsto\frac{1}{x}$ gives $\int_{1}^{+\infty}\frac{\log x}{1+x^2}\,dx = -\int_{0}^{1}\frac{\log x}{1+x^2}\,dx $.
On the other hand, $$ \int_{0}^{+\infty}\frac{\log(a)}{a^2+x^2}\,dx = \frac{\log a}{a}\int_{0}^{+\infty}\frac{dx}{1+x^2}=\color{red}{\frac{\pi\log a}{2a}}.$$
If you want to use contour integration, you may take a semicircular contour in the right half-plane with three bulges around $z=0$ (singularity for $\log z$) and $z=\pm i a$ (singularities for $\frac{1}{z^2+a^2}$), then apply the residue theorem. The integral over the outer arc is vanishing as $R\to +\infty$ since it is bounded by $\frac{2\pi\log R}{R}$ in absolute value for any $R$ big enough.