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原式=1/2∫ln(x+1)dx²
=1/2*x²ln(x+1)-1/2∫x²dln(x+1)
=1/2*x²ln(x+1)-1/2∫x²/(x+1) dx
=1/2*x²ln(x+1)-1/2∫(x²-1+1)/(x+1) dx
=1/2*x²ln(x+1)-1/2∫[x-1+1/(x+1)] dx
=1/2*x²ln(x+1)-1/4*x²+1/2x-1/2ln(x+1)+C
=1/2*x²ln(x+1)-1/2∫x²dln(x+1)
=1/2*x²ln(x+1)-1/2∫x²/(x+1) dx
=1/2*x²ln(x+1)-1/2∫(x²-1+1)/(x+1) dx
=1/2*x²ln(x+1)-1/2∫[x-1+1/(x+1)] dx
=1/2*x²ln(x+1)-1/4*x²+1/2x-1/2ln(x+1)+C
更多追问追答
追问
第四行到第五行的转换没太看懂,能详细的解答下么?
为什么第二步(x^2-1+1)到下一行里直接变成了(x-1+1)
追答
=(x²-1)/(x+1)+1/(x+1)
第一个是平方差
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