∫(2-3v)/(2v^2-3v)dv and ∫(1/1+cosu)du
pls having problem integrating the equation above, help!!!
$\displaystyle \frac{2 - 3v}{2v^2 - 3v}$
$\displaystyle = \frac{2 - 3v}{v(2v - 3)}\ =\ \frac{A}{v}\ +\ \frac{B}{2v - 3}$
(find A and B...)
$\displaystyle \frac{1}{1 + \cos u} = \frac{1 - \cos u}{(1 + \cos u)(1 - \cos u)}$
(where you have a difference of two squares on the bottom (followed by trig/pythag), and the top splits in two...)