How did the author solve this integral?

Question

I have the following spherical density distribution:

$\rho(x, z) = \frac{1}{\sqrt{x^2 + z^2}\left(1+\sqrt{x^2+z^2}\right)^2}$

which I have broken into a "line of sight" dimension $z$ and a "transverse" dimension $x$. I wish to integrate this profile along the line of sight to obtain the projected 2d-profile $\Sigma(x)$.

The published result is

(ignore the normalization prefactors)

I cannot figure out how this is obtained. For the indefinite integral, Wolfram gives

Trying to take the upper limit for the $x>1$ case via Mathematica gives

which does not simplify to the published result, as far as I can tell. Presumably there is a manual technique that Im unfamiliar with... anyone see a way to properly do this?

I should say, my end goal is to compute this integral for the general upper bound $Z$, but I wanted to first make sure I could at least reproduce the case for $Z = \infty$.

Answer 1

I observe in Mathematica (11.3)

Integrate[1/(Sqrt[x^2 + z^2] (1 + Sqrt[x^2 + z^2])^2), {z, 0, \[Infinity]}, Assumptions -> {x > 1}]

yields $$ \frac{1}{x^2-1}-\frac{\sec ^{-1}(x)}{\left(x^2-1\right)^{3/2}} \text{,} $$

Integrate[1/(Sqrt[x^2 + z^2] (1 + Sqrt[x^2 + z^2])^2) /. x -> 1, {z, 0, \[Infinity]}]

yields $$ \frac{1}{3} \text{,} $$ and

Integrate[1/(Sqrt[x^2 + z^2] (1 + Sqrt[x^2 + z^2])^2), {z, 0, \[Infinity]}, Assumptions -> {0 < x < 1}]

yields $$ -\frac{2 \sqrt{1-x^2}+\log \left(\frac{1-\sqrt{1-x^2}}{\sqrt{1-x^2}+1}\right)}{2 \left(1-x^2\right)^{3/2}} \text{.} $$

I don't necessarily recognize the identities used to complete the identification with the published result, but I do see most of the parts of the published result, so the identity of the remainders may be possible.

Further, to your intention to compute $\int_0^{\text{zz}} \dots \,\mathrm{d}z$,

Assuming[{x > 1, zz \[Element] Reals, zz > 0}, FullSimplify[ Integrate[1/(Sqrt[x^2 + z^2] (1 + Sqrt[x^2 + z^2])^2), {z, 0, zz}] ] ]

yields (for $x > 1$) $$ \frac{\text{zz} \left(\sqrt{x^2+\text{zz}^2}-1\right)}{\left(x^2-1\right) \left(x^2+\text{zz}^2-1\right)}+\frac{\tan ^{-1}\left(\frac{\text{zz}}{\sqrt{\left(x^2-1\right) \left(x^2+\text{zz}^2\right)}}\right)-\tan ^{-1}\left(\frac{\text{zz}}{\sqrt{x^2-1}}\right)}{\left(x^2-1\right)^{3/2}} $$

and I imagine the other two pieces are similar.