作业帮这个定积分怎么求啊? 求达人解惑.. ∫x^2/(1+x^2)^3dx
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这个定积分怎么求啊? 求达人解惑.. ∫<0,1>x^2/(1+x^2)^3dx
令x = tanz,dx = sec²z dz
当x = 0,z = 0；当x = 1,z = π/4
∫[0,1] x²/(1 + x²)³ dx
= ∫[0,π/4] tan²z/sec^6(z) * sec²z dz
= ∫[0,π/4] tan²z * cos^4(z) dz
= ∫[0,π/4] sin²z cos²z dz
= ∫[0,π/4] (1/2 * sin2z)² dz
= 1/4 ∫[0,π/4] sin²(2z) dz
= (1/4)(1/2) ∫[0,π/4] (1 - cos4z) dz
= 1/8 (z - 1/4 sin4z)_[0,π/4]
= (1/8)(π/4)
= π/32
当x = 0,z = 0；当x = 1,z = π/4
∫[0,1] x²/(1 + x²)³ dx
= ∫[0,π/4] tan²z/sec^6(z) * sec²z dz
= ∫[0,π/4] tan²z * cos^4(z) dz
= ∫[0,π/4] sin²z cos²z dz
= ∫[0,π/4] (1/2 * sin2z)² dz
= 1/4 ∫[0,π/4] sin²(2z) dz
= (1/4)(1/2) ∫[0,π/4] (1 - cos4z) dz
= 1/8 (z - 1/4 sin4z)_[0,π/4]
= (1/8)(π/4)
= π/32