怎么求证：∫secu(tanu)的积分
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令 x = tanu， 则 dx = (secu)^2du
I = ∫√(1+x^2)dx = ∫(secu)^3du = ∫secudtanu
= secutanu - ∫secu(tanu)^2du = secutanu - ∫secu[(secu)^2-1]du
= secutanu - I + ln|secu+tanu|
I = (1/2)[secutanu + ln|secu+tanu|] + C
= (1/2)[x√(1+x^2) + ln|x+√(1+x^2)|] + C
I = ∫√(1+x^2)dx = ∫(secu)^3du = ∫secudtanu
= secutanu - ∫secu(tanu)^2du = secutanu - ∫secu[(secu)^2-1]du
= secutanu - I + ln|secu+tanu|
I = (1/2)[secutanu + ln|secu+tanu|] + C
= (1/2)[x√(1+x^2) + ln|x+√(1+x^2)|] + C
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