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Question

xx(1 + log x) dx = ________________.

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Solution


I=xx1+logxdx

Let xx=z

logz=logxx

logz=xlogx

1zdz=x×1x+logx×1dx

dzz=1+logxdx

I=z×dzz

I=dz

I=z+C

I=xx+C


xx(1 + log x)dx = ____xx + C____.

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