# Math Help - Integral

1. ## Integral

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

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

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

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

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

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