Question

Математический анализ. Найти интеграл

Answer 1

$[u=\sqrt{x^2+1}=>du=\dfrac{2x}{2\sqrt{x^2+1}}dx=\dfrac{x}{\sqrt{x^2+1}}dx=>dx=\dfrac{udu}{\sqrt{u^2-1}}]\\ \int \dfrac{\sqrt{x^2+1}}{x}dx=\int \dfrac{u}{\sqrt{u^2-1}}\dfrac{udu}{\sqrt{u^2-1}}=\int\dfrac{u^2}{u^2-1}du=\int 1du-\int \dfrac{1}{1^2-u^2}du=u-\dfrac{1}{2}ln\dfrac{|u+1|}{|u-1|}+C=\sqrt{x^2+1}+\dfrac{1}{2}ln\dfrac{|\sqrt{x^2+1}-1|}{|\sqrt{x^2+1}+1|}+C$