令t = x + 1/2, x = t - 1/2, dx = dt, 1+x = t+ 1/2
∫√((1+x)/x)dx=∫√((1/2+t)/(t-1/2))dt
=∫(1/2+t)/√(t^2-1/4))dt
=1/2∫1/√(t^2-1/4))dt +∫t/√(t^2-1/4)dt
=1/2∫1/√(4t^2-1))d2t+1/2∫1/√(t^2-1/4)d(t^2-1/4)
令2t=secu2dt=tanu*secudu
∫1/√(4t^2-1)d2t=∫tanu*secu/√(secu^2-1) du=∫secudu=ln|secu+tanu|+C1=ln|2t+ √(4t＾2-1)|+C1
∫1/√(t^2-1/4)d(t^2-1/4)=∫(t^2-1/4)^(-1/2)d(t^2-1/4)=2(t^2-1/4)^(1/2)+C2
將t和x的關(guān)系式代入,即可求出原函數(shù).