Question

Математический анализ. Найти интеграл

Answer 1

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