<s>[INST] A faster way to evaluate $\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt$? I need to evaluate the integral
$$\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt\tag1$$
After some workarounds I found the change of variable $t=2\sqrt{x^2-1}$, then
$$\int_1^\infty\frac{2\sqrt{1+(t/2)^2}}{t^3}\,\mathrm dt=\frac12\int_{\sqrt5/2}^\infty\frac{x^2}{(x^2-1)^2}\,\mathrm dx\\=\frac12\left[\frac{x}{2(1-x^2)}\bigg|_{\sqrt5/2}^\infty+\frac12\int_{\sqrt5/2}^\infty\frac{\mathrm dx}{x^2-1}\right]\\=\frac{\sqrt5}2+\frac18\int_{\sqrt5/2}^\infty\left(\frac1{x-1}-\frac1{x+1}\right)\,\mathrm dx\\=\frac{\sqrt5}2+\frac18\ln\left(\frac{\sqrt 5+2}{\sqrt 5-2}\right)\\=\frac{\sqrt5}2+\frac14\ln(\sqrt5+2)$$
But my intuition says that it must exists a more straightforward way to evaluate this integral. In fact, using Wolfram Mathematica, I get the equivalent[*] result
$$\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt=\frac14(2\sqrt5+\operatorname{arsinh}(2))$$
[*] The equivalence can be seen from
$$\operatorname{arsinh}(x)=\ln(x+\sqrt{1+x^2})$$

My question: someone knows a faster way to evaluate manually this integral? Maybe a better change of variable?
 [/INST] Enforcing a substitution of $x \mapsto 1/x$ we have
$$I = \int_0^1 \sqrt{4 x^2 + 1} \, dx.$$
This integral can be readily found using a hyperbolic substitution of $x \mapsto \frac{1}{2} \sinh x$. Doing so yields
\begin{align*}
I &= \frac{1}{2} \int_0^{\sinh^{-1} (2)} \cosh^2 x \, dx\\
&= \frac{1}{4} \int_0^{\sinh^{-1} (2)} [\cosh (2x) + 1] \, du \tag1\\
&= \frac{1}{4} \left [\frac{1}{2} \sinh (2x) + x \right ]_0^{\sinh^{-1} (2)}\\
&= \frac{1}{8} \sinh [2\sinh^{-1} (2)] + \frac{1}{4} \sinh^{-1} (2)\\
&= \frac{\sqrt{5}}{2} + \frac{1}{4} \sinh^{-1} (2) \tag2
\end{align*}
Explanation
(1) Using $\cosh^2 x = \frac{1}{2} (\cosh (2x) + 1)$
(2) Using $\sinh (2 \alpha) = 2 \sinh \alpha \cosh \alpha$ where $\alpha = \sinh^{-1} (2)$ such that $\sinh \alpha = 2$ and $\cosh \alpha = \sqrt{5}$
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