Đặt u = ln($\sqrt{x^2+1} +x$) => du =$\frac{dx}{\sqrt{x^2+1}}$ dv = $\frac{x}{\sqrt{x^2+1}}dx$ => v = $\frac{1}{\sqrt{x^2+1}}$
I = $[\frac{1}{\sqrt{x^2+1}}ln(\sqrt{x^2+1}+x)] -\int\limits_{a}^{b}\frac{1}{x^2+1}dx$
I=............................................................- [acrtanx]