Integration of natural logarithm

Question

It is a very basic question however the sources through which I am studying are not complete hence I need a little help. How does this $$\int \frac{1+v}{1-2v-v^2}$$ Turn into this $$-\frac1 2 \ln(1-2v-v^2) $$

Answer 1

It is a standard exploitation of a more or less general rule $$ \int \frac{f'(x)}{f(x)} dx = \log x$$ Of course there are cases when the rule fails (when $f(x)$ is zero, for instance), but is an essential tool in integration.

It can be verified by evaluating the derivative, $$ \frac{d}{dx} \log f(x) $$

Answer 2

Hint. $$\int \frac{1+v}{1-2v-v^2}\text{d}v=-\frac{1}{2} \int \frac{-2-2v}{1-2v-v^2} \text{d}v$$