利用换元x=1-t/(1+t) 计算积分 0到1 ln(1+x)/(1+x^2) dx

由于0≤x=1-t/(1+t) ≤1,所以0≤t≤1,因此原积分=∫(0到1)ln[(t+1)/(t^2+1)]dt
=∫(0到1)ln(t+1)dt-∫(0到1)ln(t^2+1)dt
=tln(t+1)(t=1)-tln(t+1)(t=0)-∫(0到1)tdln(t+1)-tln(t^2+1)(t=1)+tln(t^2+1)(t=0)+∫(0到1)tdln(t^2+1)
=﹣∫(0到1)t/(t+1)×dt+∫(0到1)t/(t^2+1)×dt
=﹣∫(0到1)[1-1/(t+1)]dt+1/2∫(0到1)1/(t^2+1)×dt^2
=﹣∫(0到1)dt+∫(0到1)1/(t+1)×dt+1/2∫(0到1)1/(t^2+1)×d(t^2+1)
=﹣1+∫(0到1)1/(t+1)×d(t+1)+1/2ln(t^2+1)(t=1)-1/2ln(t^2+1)(t=0)
=﹣1+1/2ln2+ln(t+1)(t=1)-ln(t+1)(t=0)
=﹣1+3/2ln2.

以上就是解法了 “所以”其实是用方程的思想