∫1/（（1+x）（1+x²））dx 积分限为（0,+∞）怎么计算？为什么我算式中有无穷大减无穷大

∫(0->+∞) dx/[(1+x)(1+x²)]
= (1/2)∫(0->+∞) (1-x)/(1+x²) dx + (1/2)∫(0->+∞) dx/(1+x)
= (1/2)∫(0->+∞) dx/(1+x²) dx - (1/2)∫(0->+∞) x/(1+x²) dx + (1/2)∫(0->+∞) dx/(1+x)
= (1/2)arctanx - (1/4)ln(1+x²) + (1/2)|1+x|：(0->+∞)
= (1/2)arctanx - (1/4)ln(1+x²) + (1/4)ln|(1+x)²|
= (1/2)arctanx + (1/4)ln[(1+2x+x²)/(1+x²)]
= [(1/2)lim(x->+∞) arctanx - 0] + (1/4)ln[(1/x²+2/x+1)/(1/x²+1)]
= (1/2)(π/2) + (1/4)lim(x->+∞) ln[(1/x²+2/x+1)/(1/x²+1)] - (1/4)lim(x->0) ln[(1+2x+x²)/(1+x²)]
= π/4 + (1/4)ln(1) - (1/4)ln(1)
= π/4