how do u integrate 1+u/3u+2u^2
please help me get started ...tnk u
2 answers
I would do partial fractions.
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?
(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?
0 answers