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不定積分求解

Oct 17th 2013, 13:52

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(x-x ³ )^(1/3) = x (1/x ² -1)^(1/3) so ∫[(x-x ³ )^(1/3)] / x^4 dx = ∫ [(1/x ² -1)^(1/3) ]/ x ³ dx = ∫ u^(1/3) (-1/2) du ( sub. u= 1/x ² -1, du= -2/x ³ dx ) = (-3/8) u^(4/3) + C = (-3/8) (1/x ² -1)^(4/3) + C...

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