∫ [xln(x+√(1+x^2)]/(1-x^2)^2 dx 求不定积分

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∫ [xln(x+√(1+x^2)]/(1-x^2)^2 dx 求不定积分
最好写下分部积分后半部分怎么做的

∫ [xln(x+√(1+x²))]/(1-x²)² dx
=1/2∫ [ln(x+√(1+x²))]/(1-x²)² d(x²)
=-1/2∫ [ln(x+√(1+x²))]/(1-x²)² d(1-x²)
=1/2∫ [ln(x+√(1+x²))] d(1/(1-x²))
=1/2(1/(1-x²))ln(x+√(1+x²))-1/2∫ 1/(1-x²) d[ln(x+√(1+x²))]
由于：[ln(x+√(1+x²))]'=[1/(x+√(1+x²))]*(1+x/√(1+x²))=1/√(1+x²)
=1/2(1/(1-x²))ln(x+√(1+x²))-1/2∫ 1/[(1-x²)√(1+x²)] dx
令x=tanu,则√(1+x²)=secu,dx=sec²udu
=1/2(1/(1-x²))ln(x+√(1+x²))-1/2∫ 1/[(1-tan²u)secu]sec²udu
=1/2(1/(1-x²))ln(x+√(1+x²))-1/2∫ secu/(1-tan²u)du
=1/2(1/(1-x²))ln(x+√(1+x²))-1/2∫ cosu/(cos²u-sin²u)du
=1/2(1/(1-x²))ln(x+√(1+x²))-1/2∫ 1/(cos²u-sin²u)d(sinu)
=1/2(1/(1-x²))ln(x+√(1+x²))-1/2∫ 1/(1-2sin²u)d(sinu)
=1/2(1/(1-x²))ln(x+√(1+x²))+1/4∫ 1/(sin²u-0.5)d(sinu)
由公式：∫ 1/(x²-a²) dx=1/(2a)*ln|(x-a)/(x+a)|+C
=1/2(1/(1-x²))ln(x+√(1+x²))+(1/4)(1/√2)ln|(sinu-√2/2)/(sinu+√2/2)|+C
=1/2(1/(1-x²))ln(x+√(1+x²))+1/(4√2)ln|(2sinu-√2)/(2sinu+√2)|+C
由于tanu=x,则sinu=x/√(1+x²)
=1/2(1/(1-x²))ln(x+√(1+x²))+1/(4√2)ln|(2x/√(1+x²)-√2)/(2x/√(1+x²)+√2)|+C
=1/2(1/(1-x²))ln(x+√(1+x²))+1/(4√2)ln|(2x-√(2+2x²))/(2x+√(2+2x²))|+C

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