∫x[ln(1+x)]^2dx=?
1个回答
首先设t=x+1
∫x[ln(1+x)]^2dx= ∫(t-1)[lnt]^2dx
分开积分得 ∫t*[lnt]^2dx-∫[lnt]^2dx
再令lnt=u则t=e^u
代入得∫e^u*u^2de^u-∫u^2de^u
采用分部积分得
=∫0.5u^2de^2u-∫u^2de^u
=0.5(u^2*e^2u-∫e^2udu^2)-e^u*u^2+∫e^udu^2
=0.5[u^2*e^2u-(∫ude^2u)]-e^u*u^2+2∫ude^u
再次分部积分
=0.5[u^2*e^2u-(u*e^2u-∫e^2udu)]-e^u*u^2+2(u*e^u-∫e^udu)
=0.5[u^2*e^2u-(u*e^2u-0.5e^2u)]-e^u*u^2+2(u*e^u-e^u)
将u=lnt回代
得0.5[(lnt)^2*t^2-lnt*t^2+0.5t^2]-t*(lnt)^2+2t*lnt-2t
再将t=x+1代入得
积分结果为0.5[ln(x+1)]^2*(x+1)^2-0.5ln(x+1)^2+0.25(x+1)^2-(x+1)*(ln(x+1))^2+2(x+1)*ln(x+1)-2(x+1)
最后合并同类项,就好了
