Đặt\( t=\sqrt{\ln x+1} \Rightarrow t^{2} =\ln x+1\Rightarrow \ln ^{2} x=\left(t^{2} -1\right)^{2} \Rightarrow 2t{\rm d}t=\frac{1}{x} {\rm d}x.\)
Khi đó: \(\int \frac{\ln ^{2} x}{x\sqrt{\ln x+1} } {\rm d}x =\int 2\left(t^{2} -1\right)^{2} {\rm d}t \)
\(=\int \left(2t^{4} -4t^{2} +2\right){\rm d}t =2\frac{t^{5} }{5} -4\frac{t^{3} }{3} +2t+C.\)
\(=2.\frac{\left(\sqrt{\ln x+1} \right)^{5} }{5} -4.\frac{\left(\sqrt{\ln x+1} \right)^{3} }{3} +2.\sqrt{\ln x+1} +C.\)