令a=√x
x=a^2,dx=2ada
原式=∫2a/(a-2)da
2a/(a-2)=[(2a-4)+4]/(a-2)=2(a-2)/(a-2)+4/(a-2)=2+4/(a-2)
所以原式=∫2da+∫4/(a-2)da
=2a+4ln|a-2|+C
=2√x+4ln|√x-2|+C