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求定积分 1、∫(1-2){[(Inx)^2]/(x^3)}dx 2、∫(0-1){x/[e^(5x)]}dx
用分部积分法
人气:481 ℃ 时间:2020-04-27 00:54:09
解答
1.∫ { [(Inx)^2]/(x^3) } dx = (-1/2) ∫ (Inx)^2 d x^(-2)
= (-1/2) [ (Inx)^2 * x^(-2) ] + ∫ 2Inx * x^(-3) dx
= (-1/2) [ (Inx)^2 * x^(-2) ] - ∫ Inx d x^(-2)
= (-1/2) [ (Inx)^2 * x^(-2) ] - lnx ^ x^(-2) + ∫ x^(-3) dx
= (-1/2) [ (Inx)^2 * x^(-2) ] - lnx ^ x^(-2) - (1/2) x^(-2) + C
原式 = (-1/8) (ln2)^2 - ln2 /4 - 1/8
2.∫ x * e^(-5x) dx = (-1/5) ∫ x d e^(-5x) = (-1/5) x * e^(-5x) + (1/5) ∫ e^(-5x) dx
= (-1/5) x * e^(-5x) + (1/25) e^(-5x) + C
原式 = (-4/25) e^(-5) - 1/251题的第二行那里不是减号吗1. I =(-1/2) [(Inx)^2 * x^(-2)-∫2Inx * x^(-3) dx ]=(-1/2)(Inx)^2 * x^(-2)+∫ Inx * x^(-3) dx 系数改正了= (-1/2) (Inx)^2 * x^(-2)-(1/2) ∫ Inx d x^(-2)= (-1/2)(Inx)^2 * x^(-2)-(1/2) lnx ^ x^(-2) + (1/2) ∫x^(-3) dx= (-1/2)(Inx)^2 * x^(-2) -(1/2)lnx ^ x^(-2) - (1/4) x^(-2) + C原式 = (-1/8) (ln2)^2 - ln2 /8 - 1/16
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