∫▒(2-3v)/(2v^2-3v) □(24&dv

**lawochekel** · July 29th 2011, 12:58 AM

∫(2-3v)/(2v^2-3v)dv and ∫(1/1+cosu)du

pls having problem integrating the equation above, help!!!

**tom@ballooncalculus** · July 29th 2011, 02:22 AM

$\frac{2 - 3v}{2v^2 - 3v}$

$= \frac{2 - 3v}{v(2v - 3)}\ =\ \frac{A}{v}\ +\ \frac{B}{2v - 3}$

(find A and B...)

$\frac{1}{1 + \cos u} = \frac{1 - \cos u}{(1 + \cos u)(1 - \cos u)}$

(where you have a difference of two squares on the bottom (followed by trig/pythag), and the top splits in two...)

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