求教!求定积分∫(1,+∞)dx/x√1+x∧5+x∧10.
1个回答
x^10+x^5+1 = (x^5 +1/2)^2 + 3/4
let
x^5+1/2 = (√3/2)tany
5x^4 dx = (√3/2)(secy)^2 dy
x=1, y =π/3
x=+∞, y =π/2
∫(1->+∞)dx/[x√(1+x^5+x^10)]
=(1/5)∫(1->+∞)(5x^4dx)/[x^5.√(1+x^5+x^10)]
=(1/5)∫(π/3->π/2)(√3/2)(secy)^2 dy/{ [(√3/2)tany -1/2]. secy]
=(√3/5)∫(π/3->π/2)secy /(√3tany -1) dy
=(√3/5)∫(π/3->π/2)dy/(√3siny -cosy)
=(√3/10)∫(π/3->π/2)dy/[(√3/2)siny -(1/2)cosy) ]
=(√3/10)∫(π/3->π/2)dy/sin(y-π/6)
=(√3/10)∫(π/3->π/2)csc(y-π/6) dy
=(√3/10)ln|csc(y-π/6)- cot(y-π/6)| |(π/3->π/2)
=(√3/10)(ln|2/√3+1/√3| - ln|2-√3| )
=(√3/10)ln|3/(2√3-3) |
