令√x-2=t
x=t方+2
dx=2tdt
所以
原式=∫(t^2+3)/(t^2+2)t×2tdt
=2∫(t^2+3)/(t^2+2)dt
=2∫[1+1/(t^2+2)]dt
=2t +√2arctant/√2 +c
=2√(x-2)+√2arctan(√(x-2))/√2 +c