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∫1/(x^2+1)^2dx怎么算?
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∫1/(x^2+1)^2dx 怎么算?
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换元法,令x=tanu,则x²+1=sec²u,dx=sec²udu
∫1/(x^2+1)^2dx
=∫(1/sec⁴u)*sec²udu
=∫(1/sec²u)du
=∫cos²udu
=1/2∫(1+cos2u)du
=(1/2)u+(1/4)sin2u+C
=(1/2)u+(1/2)sinucosu+C
由x=tanu得：u=arctanx,sinu=x/√(1+x²),cosu=1/√(1+x²)
=(1/2)arctanx+(1/2)x/(1+x²)+C
∫1/(x^2+1)^2dx
=∫(1/sec⁴u)*sec²udu
=∫(1/sec²u)du
=∫cos²udu
=1/2∫(1+cos2u)du
=(1/2)u+(1/4)sin2u+C
=(1/2)u+(1/2)sinucosu+C
由x=tanu得：u=arctanx,sinu=x/√(1+x²),cosu=1/√(1+x²)
=(1/2)arctanx+(1/2)x/(1+x²)+C
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