\displaystyle\int\limits_1^r {{{dx} \over {8x\left( {\ln 2x} \right)^{\sqrt 2 } }}} = {{1 + \sqrt 2 } \over 8}\left( {\left( {\ln 2} \right)^{1 - \sqrt 2 } - \left( {\ln 2r} \right)^{1 - \sqrt 2 } } \right)