Prove $12\int_{0}^{\infty}{x\over (e^{x^2}+1)(e^{x^2}+3)}dx=\ln(2)$

Following Claude Leibovici's comment, assuming $\text{Re}(a),\text{Re}(b)>-1$,

$$\begin{eqnarray*}J(a,b)=\int_{0}^{+\infty}\frac{2x\,dx}{(e^{x^2}+a)(e^{x^2}+b)}&=&\int_{0}^{+\infty}\frac{du}{(e^{u}+a)(e^{u}+b)}\\&=&\frac{1}{b-a}\int_{0}^{+\infty}\left(\frac{1}{e^u+a}-\frac{1}{e^u+b}\right)\,du\\&=&\frac{1}{b-a}\int_{1}^{+\infty}\frac{dx}{x}\left(\frac{1}{x+a}-\frac{1}{x+b}\right)\\&=&\frac{1}{b-a}\int_{0}^{1}\left(\frac{1}{1+ax}-\frac{1}{1+bx}\right)\,dx\\&=&\color{red}{\frac{b\log(a+1)-a\log(b+1)}{ab(b-a)}}\end{eqnarray*}$$ and by taking the limit as $b\to a$, $$ J(a,a) = \color{red}{\frac{(a+1)\log(a+1)-a}{a^2(a+1)}}.$$