## University Calculus: Early Transcendentals (3rd Edition)

Since, we have $\int_1^\infty \dfrac{1/x}{x \sqrt {1+(\ln^2 x)}} dx$ Suppose $t=\ln x \implies dt=\dfrac{1}{x} dx$ Then $\int_1^\infty \dfrac{1/x}{x \sqrt {1+(\ln^2 x)}} dx=\int_{0}^\infty \dfrac{1}{1+u^2} dx=\lim\limits_{a \to \infty} [\tan^{-1} t]_{0}^a$ or, $=\lim\limits_{a \to \infty}[(\tan^{-1} a -\tan^{-1} (0)]$ or, $\dfrac{\pi}{2}-0$ or, $= \dfrac{\pi}{2}$ Hence, the series is convergent.