求解不定积分∫ (x^n*lnxdx)且n不等于-1

Question

如题所述

Answer 1

原式=1/(n+1)*∫ lnxdx^(n+1)
=1/(n+1)*lnx*x^(n+1)-1/(n+1)*∫x^(n+1)dlnx

=1/(n+1)*lnx*x^(n+1)-1/(n+1)*∫x^(n+1)*1/xdx
=1/(n+1)*lnx*x^(n+1)-1/(n+1)*∫x^ndx

=1/(n+1)*lnx*x^(n+1)-1/(n+1)²*x^(n+1)+C追问

怎么说呢，是我太笨了吧，能不能求一下讲解，虽然知道大概知道用分部积分法，但是还是不太懂，能讲解一下就好了，拜托了

追答

哪里不懂

追问

~\(≧▽≦)/

Answer 2

∫ (x^n*lnxdx)

=1/(n+1)∫ [lnxdx^(n+1)]
=x^(n+1)lnx/(n+1)-1/(n+1)∫x^(n+1)*1/xdx
=x^(n+1)lnx/(n+1)-1/(n+1)∫x^ndx
=x^(n+1)lnx/(n+1)-1/(n+1)^2∫dx^(n+1)
=x^(n+1)lnx/(n+1)-x^(n+1)/(n+1)^2+C