Integral

**Jason Bourne** · December 12th 2008, 08:36 AM

$\int \frac{1+v}{1-2v-v^2} dv$

I know this equals $\frac{-1}{2}log(1-2v-v^2) + C$ but I don't know how to get to it.

**Chop Suey** · December 12th 2008, 08:39 AM

Sub $u = 1 - 2v - v^2$ and see what you get.

**o_O** · December 12th 2008, 08:41 AM

${\color{red}u = 1 - 2v - v^2} \ \ \Rightarrow \ \ du = (-2 - 2v) \ dv \Leftrightarrow {\color{blue}-\frac{du}{2} = (v+ 1) \ dv}$

So: $\int \frac{{\color{blue} v+1}}{{\color{red}1-2v-v^2}} {\color{blue} \ dv} \ = \cdots$

**Jason Bourne** · December 12th 2008, 09:57 AM

thanks, I didn't realise it was that simple.

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