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求下列不定积分.
(1)∫[(1-x)²/³√x]dx;
(6) ∫[3x²/(1﹢x²)]dx;
(10) ∫[(x³-64)/(x﹣4)]dx;
人气:118 ℃ 时间:2020-06-22 15:43:01
解答
1.原式=∫(1-2x+x^2)/x^(1/3)dx=∫x^(-1/3)dx-2∫x^(2/3)dx+∫x^(5/3)dx=3/2x^(2/3)-6/5x^(5/3)+3/8x^(8/3)+C2.原式=∫(3x^2+3-3)/(1+x^2)dx=∫3dx-3∫dx/(1+x^2)=3x-3arctanx+C3.原式=∫(x^2+4x+16)dx=x^3/3+2x^2+1...
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