1∫ 1/(1 - e× ) dx
Let's add and subtract e× in the numerator
= ∫ {( 1 - e× ) + e× } / (1 - e× ) dx
= ∫ [ { (1 - e× ) / ( 1 - e× ) } + { e× / (1 - e× ) } dx ]
= ∫ 1•dx + ∫ e× / ( 1 - e× ) dx
= [ x + ∫ e× / ( 1 - e× ) dx ]. ( let's integrate second part using u substitution )
Let u = ( 1 - e× ) ∴ du = - e× dx or dx = ( - du / e× )
= [ x + ∫ - ( e× / u ) • ( 1 / e× ) • du
= [ x + ∫ - { 1/ u du }
= [ x - ∫ { 1/ u du } ]
= [ x - ln | u | + C ] { substituting back u by ( 1 - e× ) }
= [ x - ln | 1 - e× | + C ]
