Question

求不定积分：∫[(1-x^2)/(1+x^2)]^(1/2)xdx

Answer 1

∫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

Answer 2

∫[(1-x^2)/(1+x^2)]^(1/2)xdx
=(1/2)∫[(1-x^2)/(1+x^2)]^(1/2)d(x^2)
x^2=cosu sinu=√(1-x^4)
原式=(1/2)∫[(1-cosu)/(1+cosu)]^(1/2)d(cosu)
=(-1/2)∫sinutg(u/2)du=-∫[sin(u/2)]^2du=(1/2)∫(cosu-1)du=sinu/2-u/2
=[√(1-x^4)]/2-[arccos(x^2)]/2

Answer 3

x=tan(t), 标准题。