How to solve this partial fraction decomposition? Please help me to solve the following partial fraction decomposition:
$$\frac{1-v^2}{v+v^3} = \frac{A}{v}+\frac{Bv+C}{1+v^2}$$
 A: The usual method would be to multiply both sides of this equation by $v+v^3$:
$$\begin{align}\frac{1-v^2}{v+v^3} &= \frac{A}{v}+\frac{Bv+C}{1+v^2}\\
1-v^2 &= A(1+v^2) + (Bv+C)(v)\\
1-v^2 &= Av^2 + A + Bv^2 + Cv\\
-v^2 + 0v + 1 &= (A+B)v^2 + Cv + A
\end{align}$$
At this point, you can equate coefficients on the two sides. Since the $v^2$ coefficients must match, we have $-1=A+B$. Since the $v$ coefficients must match, we have $0=C$, and since the constant terms must match, we have $1=A$.
Do you see how that's working?
A: Multiply by $v $ to get
$$\frac {1-v^2}{1+v^2}=A+\frac{Bv^2+Cv}{1+v^2} $$
and then replace $v $ by $0$, to get

$$1=A $$

then multiply by $1+v^2$ to obtain
$$\frac {1-v^2}{v}=\frac {A(1+v^2)}{v}+Bv+C $$
 and replace $v $ by the complex number $i $, to find
$$\frac {2}{i}=Bi+C=-2i $$
thus

$$B=-2 \;\;,\;C=0$$

A: You want to find values for $A$, $B$, and $C$ so that 
$$\frac{A}{v}+\frac{Bv+C}{1+v^2}=\frac{1-v^2}{v+v^3}$$
or
$$A(1+v^2)+Bv^2+Cv=1-v^2$$
First, start by setting $v=0$. Then we see that
$$A(1+0)+B(0)+C(0)=1-(0)$$
$$A=1$$
Now we have
$$1(1+v^2)+Bv^2+Cv=1-v^2$$
$$Bv^2+Cv=-2v^2$$
$$Bv+C=-2v$$
Again, when we let $v=0$, we see that
$$B(0)+C=-2(0)$$
$$C=0$$
And then, of course, we see that $B=-2$ since $C=0$. Your final answer should be
$$\frac{1}{v}-\frac{2v}{1+v^2}=\frac{1-v^2}{v+v^3}$$
A: If you play around up- and downstairs you find:
\begin{align*}
\frac{1-v^2}{v + v^3} & = \frac{1-v^2 + v^2 - v^2}{v \, (1+v^2)} \\
& = \frac{1+v^2}{v(1+v^2)} + \frac{-2v^2}{v (1+v^2)}  \\
& = \frac{1}{v} - \frac{2 v}{1+v^2}
\end{align*}
A: $\begin{cases}
\displaystyle f(v)=\frac{1-v^2}{v+v^3}\\
\displaystyle g(v)=\frac{A}{v}+\frac{Bv+C}{1+v^2}\end{cases}$
With many simple partial fractions where there are not too much coefficients, it is generally possible to evaluate these coefficients by plugging some well chosen values for $v$ including equivalents in the poles and/or at infinity because they give precious information.


*

*In neighbourhood of $0$ we have $f(v)\sim \frac 1v$ and $g(v)\sim \frac Av$ thus $A=1$.

*at infinity we have $f(v)\sim -\frac 1v$ and $g(v)\sim \frac Av+\frac Bv$ thus $B=-2$.

*Finally $f(1)=0$ and $g(1)=A+\frac{B+C}2=1-1+\frac C2$ thus $C=0$

