Cute

$\begin{aligned} I & = \int_1^\infty \frac 1{x(x^2+1)}dx & \small \color{#3D99F6} \text{Let }u = x^2 \implies du = 2x \ dx \\ & = \int_1^\infty \frac 1{2u(u+1)}du \\ & = \frac 12 \int_1^\infty \left(\frac 1u - \frac 1{u+1} \right) du \\ & = \frac 12 \bigg[\ln u - \ln (u+1)\bigg]_1^\infty \\ & = \frac 12 \ln \left(\frac u{u+1}\right) \bigg|_1^\infty \\ & = \frac 12 \ln 2 \\ & = \boxed{\ln \sqrt 2} \end{aligned}$

Note first that $\dfrac{1}{x(x^{2} + 1)} = \dfrac{1 + x^{2} - x^{2}}{x(x^{2} + 1)} = \dfrac{1}{x} - \dfrac{x}{x^{2} + 1}$ .

Integrating these two terms separately, the second using the substitution $u = x^{2} + 1, du = 2x dx$ , gives us that

$\displaystyle\int \dfrac{1}{x(x^{2} + 1)} dx = \ln(x) - \dfrac{1}{2} \ln(x^{2} + 1) + C = \ln \left(\dfrac{x}{\sqrt{x^{2} + 1}}\right) + C$ .

Now $\ln \left( \dfrac{x}{\sqrt{x^{2} + 1}} \right) = \ln \left ( \dfrac{x}{x \sqrt{1 + \dfrac{1}{x^{2}}}} \right)$ , which $\to \ln(1) = 0$ as $x \to \infty$ .

Thus when we evaluate the integral as stated we obtain the answer $0 - \ln\left(\dfrac{1}{\sqrt{2}}\right) = \boxed{\ln \sqrt{2}}$ .