求定积分∫(1,0)xln(x+1)dx

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求定积分∫(1,0)xln(x+1)dx
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求定积分∫(1,0)xln(x+1)dx
求定积分∫(1,0)xln(x+1)dx

求定积分∫(1,0)xln(x+1)dx
求定积分[0,1]∫xln(x+1)dx
原式=[0,1](1/2)∫ln(x+1)dx²=[0,1](1/2){x²ln(x+1)-∫[x²/(x+1)]dx}
=[0,1](1/2){x²ln(x+1)-∫[(x-1)+1/(x+1)]dx}=[0,1](1/2){x²ln(x+1)-(x-1)²/2-ln(x+1)}
=(1/2)(1/2) =1/4

∫ x ln(x+1) dx = 1/4 (2 (x^2-1) ln(x+1)-(x-2) x)+c
从0到1积分,为1/4

∫(1,0)xln(x+1)dx
=1/2∫(1,0)ln(x+1)dx^2
=1/2x^2ln(x+1)(1,0)-1/2∫(1,0)x^2/(x+1)dx
=1/2ln2-1/2∫(1,0)[(x-1+1/(x+1)]dx
=1/2ln2-1/2[1/2x^2-x+ln(x+1)](1,0)
=1/2ln2-1/2[1/2-1+ln2]
=ln2+1/4