计算反常积分的值。。求大神详解

Question

如题所述

Answer 1

设1/[(1 + x^2)(1 + x)] = (Ax + B)/(1 + x^2) + C/(1 + x)
1 = (Ax + B)(1 + x) + C(1 + x^2)
1 = (A + C)x^2 + (A + B)x + (B + C)
则C = - A,B = - A
B + C = 1
- A - A = 1,A = - 1/2,B = C = 1/2
∫(0→∞) 1/[(1 + x^2)(1 + x)] dx
= (1/2)∫(0→∞) (- x + 1)/(1 + x^2) dx + (1/2)∫ dx/(1 + x)
= (- 1/2)∫(0→∞) x/(1 + x^2) dx + (1/2)∫(0→∞) dx/(1 + x^2) + (1/2)∫ dx/(1 + x)
= (- 1/4)ln(1 + x^2) + (1/2)arctan(x) + (1/2)ln(1 + x) |(0→∞)
= lim(x→∞) ln[√(1 + x)/(1 + x^2)^(1/4)] + (1/2)(π/2)
= lim(x→∞) ln{√[(1 + x)/x]/[(1 + x^2)/x^2]^(1/4)} + π/4
= lim(x→∞) ln[√(1 + 1/x)/(1 + 1/x^2)^(1/4)] + π/4
= ln[√(1 + 0)/(1 + 0)] + π/4
= π/4本回答被网友采纳