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求不定积分∫xln(x+1)dx
人气:295 ℃ 时间:2020-02-05 05:50:41
解答
∫xln(x+1)dx=∫ln(x+1)d(1/2*x^2)=1/2×x^2×ln(x+1)-1/2×∫x^2dln(x+1)=1/2×x^2×ln(x+1)-1/2×∫x^2/(x+1)dx=1/2×x^2×ln(x+1)-1/2×∫[x-1+1/(x+1)]dx=1/2×x^2×ln(x+1)-1/2×[1/2×x^2-x+ln(x+1...
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