Question

Решите, пожалуйста, интеграл
∫√(x²+1)dx

Answer 1

$intsqrt{x^2+1}dx=xsqrt{x^2+1}-intfrac{x^2dx}{sqrt{x^2+1}}=xsqrt{x^2+1}-intsqrt{x^2+1}dx+\+intfrac{dx}{sqrt{x^2+1}}\2intsqrt{x^2+1}dx=xsqrt{x^2+1}+ln|x+sqrt{x^2+1}|\intsqrt{x^2+1}dx=frac{1}{2}(xsqrt{x^2+1}+ln|x+sqrt{x^2+1}|)+C\\\u=sqrt{x^2+1};du=frac{xdx}{sqrt{x^2+1}}\dv=dx;v=x$