∫dx/(1+x^2)^2dx=x/(2+2x^2)+∫1/(1+x^2)dx=x/(2+2x^2)+(1/2)arctanx+c
定积分=1/2+π/4
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第1个回答 2011-12-21
∫dx/(1+x^2)^2
令x=tanu，(1+x^2)=(secu)^2，dx=(secu)^2du
原式=∫ 1/(secu)^4*(secu)^2du
=∫ (cosu)^2du
=1/2∫ (1+cos2u)du
=1/2u+1/4sin2u+C
=1/2u+1/2sinucosu+C
x=tanu，则sinu=x/√(1+x^2)，cosu=1/√(1+x^2)
=1/2arctanx+1/2*x/(1+x^2)
将上下限代入相减得：π/2+1/2
令x=tanu，(1+x^2)=(secu)^2，dx=(secu)^2du
原式=∫ 1/(secu)^4*(secu)^2du
=∫ (cosu)^2du
=1/2∫ (1+cos2u)du
=1/2u+1/4sin2u+C
=1/2u+1/2sinucosu+C
x=tanu，则sinu=x/√(1+x^2)，cosu=1/√(1+x^2)
=1/2arctanx+1/2*x/(1+x^2)
将上下限代入相减得：π/2+1/2
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