$\displaystyle \int \frac{1+v}{1-2v-v^2} dv$ I know this equals $\displaystyle \frac{-1}{2}log(1-2v-v^2) + C $ but I don't know how to get to it.
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Sub $\displaystyle u = 1 - 2v - v^2$ and see what you get.
$\displaystyle {\color{red}u = 1 - 2v - v^2} \ \ \Rightarrow \ \ du = (-2 - 2v) \ dv \Leftrightarrow {\color{blue}-\frac{du}{2} = (v+ 1) \ dv}$ So: $\displaystyle \int \frac{{\color{blue} v+1}}{{\color{red}1-2v-v^2}} {\color{blue} \ dv} \ = \cdots$
thanks, I didn't realise it was that simple.
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