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Question

Solve :
(log(logx)+1(logx)2)dx

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Solution

I=(log(logx)+1(logx)2)dx=log(logx)dx+1(logx)2dxLetI1=log(logx)dx=log(logx).1dxcontinue with integration by part:f(x)g(x)dx=f(x)g(x)dx[f(x)g(x)dx]dxnow,I1=log(logx)1dx[ddx(log(logx)).1dx]dx=xlog(logx)[1logx.1x.x]dx=xlog(logx)1logxdxLetI2=1logxdx=1logx.1dx=1logx.1dx[ddx(1logx).1dx]dx=xlogx1(logx)2.1x.xdx=xlogx+1(logx)2dxputtingthevalueofI2inI1I1=xlog(logx)xlogx1(logx)2dxnowputtingthevalueofI1inII=xlog(logx)xlogx1(logx)2dx+1(logx)2dx=xlog(logx)xlogx+C,whereC=integrationconstant

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