-2∫[1+1/(t²-1)]dt=-2∫dt-2∫1/(t²-1)dt前面半部分好算，我讲下后半部分-2∫1/(t²-1)dt=-∫[1/(t-1)-1/(t+1)]dt=-[ln(t-1)-ln(t+1)]=-ln[(t-1)/(t+1)]=+ln[(t+1)/(t-1)]分子分母同乘t+1=ln[(t+1)²/(t²-1)]=ln(t+1)²-ln(t²-1)=2ln(t+1)²-ln(t²-1)
