$\displaystyle

\int \frac{200000}{20000P - P^2} dP

$

i am so confused as to what to do, someone please help!!

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- Aug 16th 2010, 09:11 PMwik_chick88tricky integration of a fraction
$\displaystyle

\int \frac{200000}{20000P - P^2} dP

$

i am so confused as to what to do, someone please help!! - Aug 16th 2010, 09:20 PMmr fantastic
- Aug 16th 2010, 09:36 PMwik_chick88
this is what i have:

\frac{1}{P(20000-P)} = \frac{A}{P} + \frac{B}{20000-P} (i brought the 200000 out the front of the integral)

= \frac{20000A - AP + BP}{P(20000-P)}

= \frac{20000A + (B-A)P}{20000-P}

equate coefficients:

20000A = 1, therefore A = \frac{1}{20000}

B - A = 0, therefore B = A = \frac{1}{20000}

therefore:

\frac{1}{P(20000-P)} = \frac{1}{20000P} + \frac{1}{20000(20000-P)}

have i gotten it right so far? i know where to go from here, as long as i havent made any mistakes!! - Aug 16th 2010, 09:39 PMmr fantastic
- Aug 16th 2010, 09:41 PMpickslides
It is correct, math tags?

- Aug 16th 2010, 09:41 PMwik_chick88
oops forgot to wrap the text!

this is what i have:

$\displaystyle \frac{1}{P(20000-P)} = \frac{A}{P} + \frac{B}{20000-P}$ (i brought the 200000 out the front of the integral)

= $\displaystyle \frac{20000A - AP + BP}{P(20000-P)}$

= $\displaystyle \frac{20000A + (B-A)P}{P(20000-P)}$

equate coefficients:

20000A = 1, therefore A = $\displaystyle \frac{1}{20000}$

B - A = 0, therefore B = A = $\displaystyle \frac{1}{20000}$

therefore:

$\displaystyle

\frac{1}{P(20000-P)} = \frac{1}{20000P} + \frac{1}{20000(20000-P)}

$

have i gotten it right so far? i know where to go from here, as long as i havent made any mistakes!!