Consider the indefinite integral ∫(ln⁡(z))5z dz\displaystyle \int\frac{{\left({\ln{{\left({z}\right)}}}\right)}^{{5}}}{{z}}\ {\left.{d}{z}\right.}:

This can be transformed into a basic integral by letting

u=\displaystyle {u}=   and

du=\displaystyle {d}{u}=dz\displaystyle {\left.{d}{z}\right.}  

Performing the substitution yields the integral

∫\displaystyle \intdu\displaystyle {d}{u}  

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