$\int\limits\frac{\sqrt{x^{2}+1}dx}{x}$đặt $\sqrt{x^{2}+1}=u \Rightarrow x^{2}=u^{2}-1$ và $2xdx=2udu$
$\int\limits\frac{u^{2}du}{u^{2}-1}=\int\limits(1+\frac{u+1-(u-1)}{2(u-1)(u+1)})du=u+\frac{1}{2}\int\limits(\frac{1}{u-1}-\frac{1}{u+1})du$
$=u +\frac{1}{2}(ln|u-1|-ln|u+1|)+C $
oi bạn tự thay nhá !!!