Question #178d6

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Answer 1 · 2016-08-31T16:06:21

$= \frac{1}{1000} ln (\frac{x}{1000 - x}) + C$ $= 1/1000 ln ((x)/ (1000-x) )+ C$

$\int \frac{1}{x (1000 - x)} d x$ $int 1/[x(1000-x)] dx$

partial fractions is easiest/ most obvious way

$= \frac{1}{1000} \int \frac{1000}{x (1000 - x)} d x$ $=1/1000 int 1000/[x(1000-x)] dx$

$= \frac{1}{1000} \int \frac{1000 - x + x}{x (1000 - x)} d x$ $= 1/1000 int (1000-x+x)/[x(1000-x)] dx$

$= \frac{1}{1000} \int \frac{1000 - x}{x (1000 - x)} + \frac{x}{x (1000 - x)} d x$ $= 1/1000 int (1000-x)/[x(1000-x)] + (x)/[x(1000-x)] dx$

$= \frac{1}{1000} \int \frac{1}{x} + \frac{1}{1000 - x} d x$ $= 1/1000 int (1)/[x] + (1)/(1000-x) dx$

$= \frac{1}{1000} (ln x - ln (1000 - x) + C$ $= 1/1000 ( ln x - ln (1000-x) + C$

$= \frac{1}{1000} ln (\frac{x}{1000 - x}) + C$ $= 1/1000 ln ((x)/ (1000-x) )+ C$