问题描述：
高数不定积分求大侠帮忙 ∫1/[(1+x^2)^(1/2)]dx,求公式推导.结果等于ln[x+(1+x^2)^(1/2)]..
问题解答：
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∫1/[(1+x^2)^(1/2)]dx
令x=tant t=arctanx
∫1/[(1+x^2)^(1/2)]dx
=∫1/[(1+tan^2t)^(1/2)]dtant
=∫sec^2t/sectdt
=∫sectdt
=∫sect(sect+tant)/(sect+tant)dt
=∫(sec^2t+tantsect)/(sect+tant)dt
=∫1/(sect+tant)d(sect+tant)
=ln(sect+tant)+C
=ln(secarctanx+tanarctanx)+C
=ln{[(1+tan^2arctanx)^(1/2)]+x}+C
=ln[(1+x^2)^(1/2)+x]+C
令x=tant t=arctanx
∫1/[(1+x^2)^(1/2)]dx
=∫1/[(1+tan^2t)^(1/2)]dtant
=∫sec^2t/sectdt
=∫sectdt
=∫sect(sect+tant)/(sect+tant)dt
=∫(sec^2t+tantsect)/(sect+tant)dt
=∫1/(sect+tant)d(sect+tant)
=ln(sect+tant)+C
=ln(secarctanx+tanarctanx)+C
=ln{[(1+tan^2arctanx)^(1/2)]+x}+C
=ln[(1+x^2)^(1/2)+x]+C
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