高数定积分，求大神解答

Question

如题所述

Answer 1

∫(a->2ln2) dx/√(e^x-1)

let
y = √(e^x -1)
dy = {e^x/ [2√(e^x -1)] } dx
dx = [2y/(y^2+1) ] dy

∫ dx/√(e^x-1)
=2∫ dy/(y^2+1)
=2arctany + C
=2arctan√(e^x -1) + C

∫(a->2ln2) dx/√(e^x-1) =π/6
2[arctan√(e^x -1)] | (a->2ln2) =π/6
2[ π/3 - arctan√(e^a -1) ] =π/6
arctan√(e^a -1) = π/4
√(e^a -1) =1
e^a -1 =1
a = ln2

(2)
∫(0->∞) x^7 .e^(-x^2) dx
=(-1/2)∫(0->∞) x^6 de^(-x^2)
=(-1/2)[x^6.e^(-x^2)] |(0->∞) +3∫(0->∞) x^5 e^(-x^2) dx
=3∫(0->∞) x^5 e^(-x^2) dx
=(-3/2)∫(0->∞) x^4 d.e^(-x^2)
=(-3/2)[x^4.e^(-x^2)] |(0->∞) +6∫(0->∞) x^3 .e^(-x^2) dx
=-3∫(0->∞) x^2 .de^(-x^2)
=-3[x^2.e^(-x^2)] |(0->∞) + 6∫(0->∞) x .e^(-x^2) dx
=-3[e^(-x^2)]|(0->∞)
=3