What am I doing wrong? Partial Fraction Decomp. $$\int\frac{1-v}{(v+1)^2}dv$$

I think I am supposed to do PFD, but solving for A and B I get zero for both.

$$(1-v) = A(v+1) + B(v+1)$$

let $v = -1$

$$A = \frac{2}{0}, B = \frac{2}{0}$$

So this is undefined? (or infinity?)


If you’re interested in the integration rather than the abstract partial fraction problem, you should make the substitution $u=v+1$, giving
$$
\int\frac{(2-u)du}{u^2}=\int\Big(2u^{-2}-\frac1u\Big)du\,,
$$
then make the backward substitution after doing your simple integration.
