A better approximation than mine for $\int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx$, where $\operatorname{gd}(x)$ is the Gudermannian function Let $\operatorname{gd}(x)$ the Gudermannian function, defined as in this MathWorld's article, and implemented in Wolfram Language as Gudermannian[x]. 
This afternoon I've spent an hour playing with Wolfram Alpha online calculator about integrals, for instance, like this $$I=\int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx.$$
Notice that the plot in previous article of MathWorld tell us that for $0<x<1$ one has $0<\operatorname{gd}(x)<1$. 
My calculations to get an approximation were $$I\approx\int_0^1 \operatorname{gd}(x)\,dx=-\frac{\pi}{2}+i\left(\operatorname{Li_2}(-ie)-\operatorname{Li_2}(ie)\right)-2C,$$ where I've combined with the indefinite integral of the Gudermannian functions that tell us MathWorld. Here $C$ is Catalan's constant. But this quantity is about $\approx 0.464065$.
But WA calculated $I$ as $0.365619$, play this code: 
integrate log(1+(gd(x)))dx, from x=0 to x=1

Question. Can you provide us a better approximation, than mine, for our integral $$\int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx?$$
  Please provide also the more important justifications in your calculations. Also if your approach uses the numerical analysis and/or calculations with your computer provide us some detail of your calculation/method. Thanks in advance.

 A: The Gudermannian function is lower-bounded by the line going to $(0,0)$ and $(1,\operatorname{gd}(1))$. However, it doesn't make a very large difference, and it hence gives a reasonable approximation. 
See, for example, a plot here on Wolfram Alpha.
Now note that 
\begin{align*} \int_0^1\log\left(1+\operatorname{gd}(x)\right)\,dx &> \int_0^1\log\left(1+\operatorname{gd}(1)x\right)\,dx \\ &= \left[ x \log(\operatorname{gd}(1) x + 1) + \frac{\log(\operatorname{gd}(1) x + 1)}{\operatorname{gd}(1)} - x\right]^1_0 \\ & =  \log(\operatorname{gd}(1)+ 1)  +\frac{\log(\operatorname{gd}(1)+ 1)}{\operatorname{gd}(1) }  - 1 \\ &\approx 0.344043 \end{align*}

Even more simply, one can approximate the integral by $$\frac{\log(1+\operatorname{gd}(1))+\log(1+\operatorname{gd}(0))}{2} = \frac{\log(1+\operatorname{gd}(1))}{2} \approx 0.31184$$
A: You considered $\log(1+x)\approx x$ forgetting that gd(x) goes from 0 to 0.6 circa on the given interval, that is too much and gives a huge error
Compare $\log(1+0.4)\approx 0.33;\;\log(1+0.5)\approx 0.4$ and $\log(1+0.6)\approx 0.47$ 
This causes your approximation being too high. Taking  integration limit lower than 1, like 0.2 or 0.3 gives good approximation.
