# Prove that $\int_{0}^{\infty}{\operatorname{arccot}^2(\sqrt{1+x})\over 1+x}\mathrm dx=\pi C-{7\over 4}\zeta(3)$

$$I=\int_{0}^{\infty}{\operatorname{arccot}^2(\sqrt{1+x})\over 1+x}\mathrm dx$$

$$J=4\int_{0}^{\infty}{\operatorname{arccot}^2[(1+x)^2 ]\over 1+x}\mathrm dx$$

Q1: Show that $I=J\color{red}?$

Q2: How can we prove that $$I=\pi C-{7\over 4}\zeta(3)\color{red}?$$

Where $C$ is Catalan's constant

$u=\sqrt{1+x}\implies 2udu=dx$

$$I=2\int_{1}^{\infty}{\operatorname{arccot}^2(u)\over u}\mathrm du$$

$v=(1+x)^2\implies dv=2(1+x)dx$

$$J=2\int_{1}^{\infty}{\operatorname{arccot}^2(v)\over v}\mathrm dv$$

$\ds{\color{#f00}{\large\S}}$: See Catalan Constant Page.
You have already showed that $I=J$: they are associated with the same integral, i.e.
$$I=2\int_{0}^{1}\frac{\arctan^2(u)}{u}\,du = 4 \int_{0}^{\pi/4}\frac{t^2}{\sin(2t)}\,dt=\frac{1}{2}\int_{0}^{\pi/2}\frac{z^2}{\sin z}\,dz\tag{1}$$ The claim now follows from the Eisenstein series for $\frac{1}{\sin z}$:
$$\frac{1}{\sin z} = \frac{1}{z}+\sum_{n\geq 1}(-1)^n\left(\frac{1}{z-n\pi}+\frac{1}{z+n\pi}\right) \tag{2}$$ by multiplying both sides by $z^2$, performing integration over $(0,\pi/2)$ and rearranging.
An efficient alternative is to use a step of integration by parts to get $I=\int_{0}^{\pi/2}z\log\tan\frac{z}{2}\,dz$, then exploit the Fourier series of $\log\sin$ and $\log\cos$.