求不定积分∫xln(x+1)dx

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优质解答：

∫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)]＋C

＝1/2×(x^2－1)×ln(x+1)－1/4×(x^2－2x)＋C

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