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integral of log x
 Let integral(logxdx) = I

integration by parts. Let logx = u, dx = dv

v = integral(dv) = x

du = d(log x)/d x = 1/x

Then int(u dv) = u v - int(v du)

So int(log x dx) = x logx - int(x.dx/x)
 
Hence int(logxdx) = xlogx - int(dx)

Therefore I = xlogx - x = x(logx - 1) + C


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