求定积分dx/(1+√(1+x^2)),
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注:tan(1/2*α)=(sin α)/(1+cos α)=(1-cos α)/sin α
令x=tant
则∫dx/(1+√(1+x^2))
=∫((sect)^2/(1+sect))dt
=∫(1/(cost(1+cost)))dt
=∫(∫1/cost - 1/(1+cost))dt
=ln|sect+tant| -1/2∫(1/(cos(t/2))^2)dt
=ln|sect+tant| -tan(t/2)+C
=ln|√(1+x^2) + x| -(√(1+x^2)-1)/x +C
令x=tant
则∫dx/(1+√(1+x^2))
=∫((sect)^2/(1+sect))dt
=∫(1/(cost(1+cost)))dt
=∫(∫1/cost - 1/(1+cost))dt
=ln|sect+tant| -1/2∫(1/(cos(t/2))^2)dt
=ln|sect+tant| -tan(t/2)+C
=ln|√(1+x^2) + x| -(√(1+x^2)-1)/x +C
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