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求不定积分∫xln(x+1)dx

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求不定积分∫xln(x+1)dx
求不定积分∫xln(x+1)dx
∫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