∫ [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+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+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+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

∫ x ln[(x+√(1+x²)] / (1-x²)² dx = (1/2) ∫ ln[(x+√(1+x²)] d [1/(1-x²)]
= (1/2) ln[(x+√(1+x²)] /(1-x²) ﹣(1/2) ∫ 1/(1-x²) /√(1+x²) dx ①

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∫ x ln[(x+√(1+x²)] / (1-x²)² dx = (1/2) ∫ ln[(x+√(1+x²)] d [1/(1-x²)]
= (1/2) ln[(x+√(1+x²)] /(1-x²) ﹣(1/2) ∫ 1/(1-x²) /√(1+x²) dx ①
其中 I2 = ∫ 1/(1-x²) /√(1+x²) dx 利用换元法求。
令 x = tant, dx = sec²t dt
I2 = ∫ sect /(1﹣tan²t) dt = ∫ cost / (cos²t - sin²t) dt
= ∫ cost / (1 ﹣2sin²t) dt 令 u= sint
= (1/2) ∫ du / (1/2﹣u²) = (√2/4) ln |(√2/2+u) /(√2/2﹣u)| + C
= (√2/4) ln| (1+√2u) /(1﹣√2u)| + C u= sint = x /√(1+x²)
= (√2/4) ln| (√(1+x²)+√2x) /(√(1+x²)﹣√2x)| + C
代入 ① 中即可。

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