√(3x-2)/x的不定积分

令t=√(3x-2),得x=(t^2+2)/2,dx=t dt
∫√(3x-2)/xdx
=∫2t/(t^2+2)·t dt
=2∫(t^2)/(t^2+2) dt
=2∫(t^2+2-2)/(t^2+2) dt
=2[∫1dt-2∫1/(t^2+2)]dt
=2∫1dt-4√2∫1/[(t/√2)^2+1]d(t/√2)
=2t-4√2 arctan(t/√2)+C
=2√(3x-2)-4√2arctan√(3x/2-1)+C