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.. ∫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] sin&...