求不定积分：∫[(1-x^2)/(1+x^2)]^(1/2)xdx
3个回答

∫x√[(1-x²)/(1+x²)] dx

设u=x²,du=2xdx

=(1/2)∫√[(1-u)/(1+u)] du

设t=√[(1-u)/(1+u)

u=(1-t²)/(1+t²)

du=-4t/(1+t²)² dt

=-2∫t²/(1+t²)² dt

令t=tanθ,dt=sec²θdθ

(1+t²)²=sec⁴θ

=-2∫tan²θ/sec⁴θ*sec²θ dθ

=-2∫tan²θ/sec²θ dθ

=-2∫(sec²θ-1)/sec²θ dθ

=-2∫dθ+2∫cos²θ dθ

=-2θ+∫(1+cos2θ)dθ

=-θ+sinθcosθ+C

=-arctan(t)+t/(1+t²)+C

=-arctan√[(1-u)/(1+u)]+(1/2)√(1-u²)+C

=-arctan√[(1-x²)/(1+x²)]+(1/2)√(1-x⁴)+C

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