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January 31, 2015

January 31, 2015

Posted by **allison** on Wednesday, October 8, 2008 at 7:22pm.

please help me get started ...tnk u

- plz integrate -
**bobpursley**, Wednesday, October 8, 2008 at 7:24pmI would do partial fractions.

- plz integrate -
**Damon**, Wednesday, October 8, 2008 at 7:55pmassume you mean

(1+u) du /(3u+2u^2)

which is

(1+u) du/[u(3+2u)]

which is

1 du/[u(3+2u)] + u du/[u(3+2u)]

now let's see if we can expand 1/[u(3+2u)]

into

a/u + b/(3+2u) = 1/ [u(3+2u)]

a(3+2u) + b(u) = 1

3 a = 1 so a = 1/3

2a + b = 0 so b = -2/3

so we have

1/[u(3+2u)] = (1/3)/u - (2/3)/(3+2u)

so the integral now is

(1/3)du/u -(2/3)du/(3+2du) +du/(3+2u)

or

(1/3)du/u +(1/3) du/(3+2u)

ok from there?

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