A difficult improper integral 
What is the value of the $$\int_{1}^{\infty}\frac{dx}{ x^2(e^x+1)}?$$

I got  $2$ as a result.  Wolfram has given $0.11111$ as a result. I am confused.
Is it correct? Or incorrect ? Please tell me...
Thanks in advance

We are dealing with
$$ \int_{1}^{+\infty}\frac{1-\tanh(x/2)}{2x^2}\,dx=\frac{1}{2}-\frac{1}{4}\int_{1/2}^{+\infty}\frac{\tanh(z)}{z^2}\,dz=\frac{1}{2}-\int_{1/2}^{+\infty}\sum_{n\geq 0}\frac{2\,dz}{z(4z^2+(2n+1)^2\pi^2)}$$
or
$$ \frac{1}{2}-\sum_{n\geq 0}\frac{\log(1+(2n+1)^2 \pi^2)}{(2n+1)^2 \pi^2}=\frac{1}{2}+\frac{\gamma}{4}+\frac{\log 2}{3}-3\log A-\sum_{n\geq 0}\frac{\log\left(1+\frac{1}{(2n+1)^2\pi^2}\right)}{(2n+1)^2 \pi^2} $$
where the last series can be numerically approximated by exploiting the Padé relation $\log(1+z^2)\approx\frac{z^2}{1+\frac{z^2}{2}}$ in a neighborhood of the origin. This leads to
$$ \int_{1}^{+\infty}\frac{dx}{x^2(e^x+1)}\,dx\approx \frac{1}{12} \left(3+3\gamma+4\log 2-36\log A+6 \sqrt{2}\tanh\frac{1}{2\sqrt{2}}\right)=\color{green}{0.11916}9\ldots$$
