∫ dx/(x≤ft(x^2+1)) equals

$I = \int{\frac{dx}{x(x^2+ 1)}}=\int\frac{x^{-3}}{1+x^{-2}}dx$

$1+x^{-2}=y$

$dy=-2x^{-3}dx$

so $I=-\frac{1}{2}\int\frac{dy}{y}=-\frac{1}{2}\log(y)+C$

$=-\frac{1}{2}\log(1+x^{-2})+C$

$=\log x-\frac{1}{2}\log(x^2+1)+C$