What is the value of the following definite integral? $I=\displaystyle\int\limits_0^\infty \dfrac{1}{(1+x^2)\displaystyle\sqrt{\log(1+x)}}dx$
My attempt:
Substituting $x=\tan{\theta}$
$I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{1}{\displaystyle\sqrt{\log(1+\tan\theta)}}d\theta$
By applying $\,\,\int\limits_a^b f(x)dx=\int\limits_a^b f(a+b-x)dx$
$I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{1}{\displaystyle\sqrt{\log(1+\cot\theta)}}d\theta$
Adding both and simplifying,
$2I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{\displaystyle\sqrt{\log(1+\cot\theta)}+\displaystyle\sqrt{\log(1+\tan\theta)}}{\displaystyle\sqrt{\log(1+\cot\theta)\log(1+\tan\theta)}}d\theta$
Rationalizing the numerator,
$2I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{\log(1+\cot\theta)-\log(1+\tan\theta)}{\displaystyle\sqrt{\log(1+\cot\theta)\log(1+\tan\theta)}\displaystyle\left(\sqrt{\log(1+\cot\theta)}-\displaystyle\sqrt{\log(1+\tan\theta)}\right)}d\theta$
$\therefore\, 2I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{\log(1+\tan\theta)-\log(1+\tan\theta)-\log\tan\theta}{\displaystyle\sqrt{\log(1+\cot\theta)\log(1+\tan\theta)}\displaystyle\left(\sqrt{\log(1+\cot\theta)}-\displaystyle\sqrt{\log(1+\tan\theta)}\right)}d\theta$
$\therefore\, 2I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{-\log\tan\theta}{\displaystyle\sqrt{\log(1+\cot\theta)\log(1+\tan\theta)}\displaystyle\left(\sqrt{\log(1+\cot\theta)}-\displaystyle\sqrt{\log(1+\tan\theta)}\right)}d\theta$
Again applying the property $\,\,\int\limits_a^b f(x)dx=\int\limits_a^b f(a+b-x)dx$
$\therefore\, 2I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{-\log\cot\theta}{\displaystyle\sqrt{\log(1+\tan\theta)\log(1+\cot\theta)}\displaystyle\left(\sqrt{\log(1+\tan\theta)}-\displaystyle\sqrt{\log(1+\cot\theta)}\right)}d\theta$
Adding both equations,
$ 2I=\displaystyle\int\limits_0^{\frac{\pi}{2}} \dfrac{-\log\tan\theta+\log\tan\theta}{\displaystyle\sqrt{\log(1+\tan\theta)\log(1+\cot\theta)}\displaystyle\left(\sqrt{\log(1+\tan\theta)}-\displaystyle\sqrt{\log(1+\cot\theta)}\right)}d\theta$
$\therefore \, 4I=0$
$\therefore \, I=0$
But DESMOS gives me the answer approaching $2.53824756838$
What wrong with my approach?
 A: Doubt this integral has an elementary form, however it can be expressed as a series:
$$\int_0^\infty \frac{dx}{(1+x^2) \sqrt{\ln(1+x)}}=\int_0^\infty \frac{e^u du}{(1+(e^u-1)^2) \sqrt{u}}=2\int_0^\infty \frac{e^{v^2} dv}{2e^{-v^2}-2+e^{v^2}}$$
We work with the last integral:
$$2\int_0^\infty \frac{e^{v^2} dv}{2e^{-v^2}-2+e^{v^2}}=2\int_0^\infty \frac{e^{-v^2}dv}{1-2e^{-v^2}(1-e^{-v^2})}$$
It's easy to check that $|2e^{-v^2}(1-e^{-v^2})|<1$, so we can use the geometric series formula:
$$2\int_0^\infty \frac{e^{-v^2}dv}{1-2e^{-v^2}(1-e^{-v^2})}=2\sum_{n=0}^\infty 2^n \int_0^\infty e^{-(n+1)v^2}(1-e^{-v^2})^n dv=$$
Now we use the binomial sum formula:
$$=2\sum_{n=0}^\infty 2^n \sum_{k=0}^n (-1)^k \binom{n}{k}  \int_0^\infty e^{-(n+k+1)v^2}dv=\sqrt{\pi} \sum_{n=0}^\infty 2^n \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{\sqrt{n+k+1}}$$
For the last part we used the well known Poisson integral formula. Finally:

$$\int_0^\infty \frac{dx}{(1+x^2) \sqrt{\ln(1+x)}}=\sqrt{\pi} \sum_{n=0}^\infty 2^n \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{\sqrt{n+k+1}}$$

It is easy to check numerically with Mathematica that the series gives the same value as the integral.
At a first glance, the convergence of this series is doubtful. I will not prove it here, but numerically with Mathematica it's clear that the series converges, and fast (here's the sum plotted vs the number of terms):

