Răspuns:
A.
Explicație pas cu pas:
[tex]\displaystyle \int_{1}^x\dfrac{1}{t(t^2+1)}\, \mathrm{dt} =\dfrac{1}{2}\int_{1}^x\dfrac{2t}{t^2(t^2+1)}\ \mathrm{dt} =\\ \\ \\ t^2+1 = u \Rightarrow 2t\, \mathrm{dt} = \mathrm{du} \\ t = 1 \Rightarrow u = 2,~~t = x \Rightarrow u = x^2+1 \\ t^2 = u-1\\ \\ = \dfrac{1}{2}\int_{2}^{x^2+1}\dfrac{1}{(u-1)u}\, \mathrm{du} = \int_{2}^{x^2+1}\dfrac{u-(u-1)}{(u-1)u}\, \mathrm{du}=\\ \\ = \dfrac{1} {2}\Big(\int_{2}^{x^2+1}\dfrac{1}{u-1}\, \mathrm{du} - \int_{2}^{x^2+1}\dfrac{1}{u}\,\mathrm{du}\Big)=[/tex]
[tex]= \dfrac{1}{2}\ln\Big|u-1\Big|\Bigg|_{2}^{x^2+1}- \dfrac{1}{2}\ln |u|\Big|_{2}^{x^2+1} = \\ \\ = \dfrac{1}{2}\Big(\ln(x^2)-\ln(x^2+1)+\ln(2)\Big) = \dfrac{1}{2}\ln\Big(\dfrac{2x^2}{x^2+1}\Big) \\ \\ \Rightarrow l = \lim\limits_{x\to \infty}\dfrac{1}{2}\ln\Big(\dfrac{2x^2}{x^2+1}\Big) = \dfrac{1}{2}\ln \Big(\lim\limits_{x\to \infty}\dfrac{2x^2}{x^2+1}\Big)=\\ \\ = \dfrac{1}{2}\ln 2 = \boxed{\ln \sqrt 2}[/tex]
