$I=\int\limits\frac{dx}{\sqrt{4x^2-4x+2}}=\int\limits\frac{dx}{\sqrt{(2x-1)^2+1}}$Đặt $2x-1=\tan t\Rightarrow dx=\frac{dt}{2\cos^2t}$$\Rightarrow I=\frac{1}{2}\int\limits\frac{dt}{\cos^2t\sqrt{\tan^2t+1}}=\frac{1}{2}\int\limits\frac{dt}{\cos^2t\sqrt{\frac{1}{\cos^2t}}}=\frac{1}{2}\int\limits\frac{dt}{\cos t}$$=\frac{1}{2}\int\limits\frac{\cos tdt}{1-\sin^2t}$$=-\frac{1}{2}\int\limits\frac{d(sint)}{sin^2t-1}$$=-\frac{1}{2}\int\limits(\frac{1}{sint-1}-\frac{1}{sint+1})d(sint)$$=\frac{1}{2}\int\limits(\frac{1}{sint+1}-\frac{1}{sint-1})d(sint)$$=\frac{1}{2}ln\left| {\frac{sint+1}{sint-1}} \right|+C$

$I=\int\limits\frac{dx}{\sqrt{4x^2-4x+2}}=\int\limits\frac{dx}{\sqrt{(2x-1)^2+1}}$Đặt $2x-1=\tan t\Rightarrow dx=\frac{dt}{2\cos^2t}$$\Rightarrow I=\frac{1}{2}\int\limits\frac{dt}{\cos^2t\sqrt{\tan^2t+1}}=\frac{1}{2}\int\limits\frac{dt}{\cos^2t\sqrt{\frac{1}{\cos^2t}}}=\frac{1}{2}\int\limits\frac{dt}{\cos t}$$=\frac{1}{2}\int\limits\frac{\cos tdt}{1-\sin^2t}=\frac{1}{2}\int\limits\frac{d(\sin x)}{1-\sin^2x}$$=\frac{1}{4}\int\limits(\frac{1}{1-\sin x}+\frac{1}{1+\sin x})d(\sin x)$$=\frac{1}{4}(-\ln|{1-\sin x}|+\ln|1+\sin x|)=\frac{1}{4}\ln|\frac{1+\sin x}{1-\sin x}|+C$