First slide
Methods of integration
Question

dxx(logx)2+4logx+1

Moderate
Solution

dxxlogx2+4logx+1

 Put logx=t

1xdx=dtdxx(logx)2+4logx+1=dtt2+4t+1

 Use 1x2a2dx=12alogxax+a+c

=dtt2+2.2t+(2)2+1(2)2=dt(t+2)2(3)=dt(t+2)2(3)2=123logt+23t+2+3+c=123loglogx+23logx+2+3

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