how are we supposed to find the integration of

log x upon {(1+log x) the whole square}

Let logx =t,then e^t=x =>e^tdt=dx then intg becomes te^tdt/(1+t)^2 now use the formula =>§e^x( f(x)+f'(x))dx=e^xf(x).where §=integration.

Put log x + 1 = t => log x = t - 1 => dx = e^t dt answer is [ x / ( log x + 1 )]