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# Solution to a integration problem

B.Sudhakar
16/01/2017 0 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 ]

0 Dislike

K

Karthik | 17/01/2017

Thanks sumeru please keep posting more and more lessons to help students.

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