Question #178d6

Question

1393 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-31T16:06:21

#= 1/1000 ln ((x)/ (1000-x) )+ C #

#int 1/[x(1000-x)] dx #

partial fractions is easiest/ most obvious way

#=1/1000 int 1000/[x(1000-x)] dx #

#= 1/1000 int (1000-x+x)/[x(1000-x)] dx #

#= 1/1000 int (1000-x)/[x(1000-x)] + (x)/[x(1000-x)] dx #

#= 1/1000 int (1)/[x] + (1)/(1000-x) dx #

#= 1/1000 ( ln x - ln (1000-x) + C #

#= 1/1000 ln ((x)/ (1000-x) )+ C #