Question

求-1到1的定积分 ∫√(1+(2x/x^2-1)^2)dx

Answer 1

I = ∫ [-1,1] (1+x²) /(1-x²) dx = 2 ∫ [0,1] (1+x²) /(1-x²) dx
= 2 ∫ [0,1] [ -1 + 2 /(1-x²) ] dx 瑕积分， 瑕点 x=1
= 2 [ - x + ln(1+x)/(1-x) ] | [0,1]
发散