$M =\displaystyle\int\dfrac{1}{x\sqrt{x+1}}dx$
Đặt $u = \sqrt{x+1}$
$\to du =\dfrac{1}{2\sqrt{x+1}}dx$
Ta được:
$M = 2\displaystyle\int\dfrac{1}{u^2 -1}du$
$\to M =-2\displaystyle\int\left(\dfrac{1}{2(u+1)} -\dfrac{1}{2(u-1)}\right)dx$
$\to M = -\displaystyle\int\dfrac{1}{u+1}du +\displaystyle\int\dfrac{1}{u-1}du$
$\to M = -\ln|u+1| + \ln|u-1| + C$
$\to M =\ln|\sqrt{x+1} -1| - \ln|\sqrt{x+1} +1| + C$
