Question

求数学大神，求解不定积分

Answer 1

let
(x+1)^(1/6) = tanu
(1/6)(x+1)^(-5/6) dx = (secu)^2 du
dx = 6(tanu)^5. (secu)^2 du
∫ dx/[ 1+ (x+1)^(1/3) ]
=∫ 6(tanu)^5. (secu)^2 du/ (secu)^2
=6∫ (tanu)^5 du
=6∫ (tanu)^3. [ (secu)^2 -1] du
=6∫ (tanu)^3 dtanu - 6∫ (tanu)^3 du
=(3/2)(tanu)^4 -6∫ tanu .[ (secu)^2 -1] du
=(3/2)(tanu)^4 - 3(tanu)^2 +6∫ tanu du
=(3/2)(tanu)^4 - 3(tanu)^2 -6ln|cosu| +C
=(3/2)(x+1)^(2/3) - 3(x+1)^(1/3) -6ln|1/√[1+ (x+1)^(1/3)] | +C
=(3/2)(x+1)^(2/3) - 3(x+1)^(1/3) +3ln|1+ (x+1)^(1/3) | +C

Answer 2

可以考虑换元法，答案如图所示

Answer 3

换元法。

看过程体会

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