Last Activity: 13 Years ago
I = xnlogx dx
choose logx as first function & xn as second then using integration by parts
I = logx [xndx] - { d/dx (logx) [xndx]}dx
I = logx[xn+1]/n+1 - { xn+1/x(n+1)}dx
I = logx[xn+1]/n+1 - { xn/n+1}dx
I = logx[xn+1]/n+1 - {xn+1}/(n+1)2 + C
approve if u like my ans
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Last Activity: 2 Year ago(s)