Posted by allison on Wednesday, October 8, 2008 at 7:22pm.
how do u integrate 1+u/3u+2u^2
please help me get started ...tnk u

plz integrate  bobpursley, Wednesday, October 8, 2008 at 7:24pm
I would do partial fractions.

plz integrate  Damon, Wednesday, October 8, 2008 at 7:55pm
assume 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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