令 x e^x = u,du = (x+1) e^x dx∫ (x+1) / [x (1+ x e^x)] dx = ∫ (x+1) e^x / [ x e^x (1+ x e^x)] dx = ∫ du / [ u (u+1)] = ln | u/(u+1)| + C= ln | x e^x / (x e^x +1) | + C
