# Integral ∫∞0arccot(√x−2√x+1)x+1dx\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx

Is it possible to evaluate this integral in a closed form?

Define the function $I(s)$ for $s > 0$ by

By observing that $I(\infty) = 0$, we have

Applying Leibniz’s integral rule,

Now with the substitution $x = \tan^{2}\theta$,

Then with the substitution $z = e^{i\theta}$, it follows that

where the contour $\Gamma$ denotes the counter-clockwise semicircular arc joining from $-i$ to $i$. Note that we have $z^{2} - 2z \cosh s + 1 = 0$ if and only if $\cosh s = \cos \theta$ if and only if $z = e^{i\theta} = e^{\pm s}$. Thus deforming the contour to the straight line joining from $-i$ to $i$ with infinitesimal indent at the origin, we obtain

where $f$ denotes the integrand

Proceeding the calculation,

Now with the substitution $u = \frac{1}{2} \{ x + x^{-1} \}$, it follows that

Finally, with the substitution $x = \sinh t$,

Our problem corresponds to $s = \log 2$, or equivalently, $a := \sinh s = \frac{3}{4}$.

where the function

is the inverse tangent integral. Using the following simple identity

it follows that

Therefore, plugging $a = \frac{3}{4}$ gives

as Cleo pointed out without explanation. More generally, for $k > 1$