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Aus Online Mathematik Brückenkurs 2

It is simpler to investigate the integral if we write it as

lnxx1dx

The derivative of lnx is 1x, so if we choose u=lnx, the integral can be expressed as

uudx.

Thus, it seems that u=lnx is a useful substitution,

lnxx1dx=udu=lnx=(lnx)dx=(1x)dx=udu=21u2+C=21(lnx)2+C.