Converting to partial fractions I would like to see a step by step conversion of this fraction to two partial fractions.
$\frac{x^2}{1-x^2}$

As André said, because the degree of the numerator is greater than or equal to that of the denominator, the first step is to divide out to get polynomial quotient and a rational function remainder whose numerator has smaller degree than its denominator:
$$\frac{x^2}{1-x^2}=-1+\frac1{1-x^2}\;.$$
Now expand the second term. The denominator factors as $(1-x)(1+x)$, with no repeated factors, so you expect to be able to decompose the fraction as 
$$\frac1{1-x^2}=\frac{A}{1-x}+\frac{B}{1+x}\tag{1}$$
for some constants $A$ and $B$. Now recombine the righthand side of $(1)$:
$$\begin{align*}
\frac1{1-x^2}&=\frac{A}{1-x}+\frac{B}{1+x}\\
&=\frac{A(1+x)+B(1-x)}{1-x^2}\\
&=\frac{(A-B)x+(A+B)}{1-x^2}\;.
\end{align*}\tag{2}$$
The left and right sides of $(2)$ are fractions with the same denominator, and they’re supposed to be the same function, so we must choose $A$ and $B$ in such a way that they have the same numerator:
$$1=(A-B)x+(A+B)\;.\tag{3}$$
Two polynomials are identically equal (i.e., equal for all values of the variable) if and only if they have identical coefficients, so the next step is to equate coefficients. On the lefthand side of $(3)$ the coefficient of $x$ is $0$, and the constant term is $1$; on the righthand side the coefficient of $x$ is $A-B$, and the constant term is $A+B$. Thus, we want $A$ and $B$ to satisfy the system
$$\left\{\begin{align*}
A-B&=0\\
A+B&=1
\end{align*}\right.$$
This system is easily solved to yield $A=B=1/2$, which we can insert into $(1)$ to get $$\frac1{1-x^2}=\frac12\left(\frac1{1-x}+\frac1{1+x}\right)$$ and, finally, $$\frac1{1-x^2}=-1+\frac12\left(\frac1{1-x}+\frac1{1+x}\right)\;.$$
Several shortcuts are possible, but this is the procedure that they’re cutting short.
