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Question

Integrate the function x(logx)2

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Solution

I=x(logx)2dx
Taking (logx)2 as first function and 1 as second function and integrating by part, we obtain
I=(logx)2xdx[{(ddxlogx)2}xdx]dx
=x22(logx)2[2logx1xx22dx]
=x22(logx)2xlogxdx
Again integrating by parts, we obtain
I=x22(logx)2[logxxdx{(ddxlogx)xdx}dx]
=x22(logx)2[x22logx1xx2xdx]
=x22(logx)2x22logx+12xdx
=x22(logx)2x22logx+x24+C

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