Why is: $\frac{x^2}{(x^2-1)(x^2+1)} = \frac{1}{2}\left(\frac{1}{x^2-1}\right)+\frac{1}{2}\left(\frac{1}{x^2+1}\right)$

Question

I am trying to understand the following.

$\dfrac{x^2}{(x^2-1)(x^2+1)} = \dfrac{1}{2}\left(\dfrac{1}{x^2-1}\right)+\dfrac{1}{2}\left(\dfrac{1}{x^2+1}\right)$

If I start with the right side I can easily get to the left side of the equation but not the other way around

Answer 1

Let $y:=x^2$. Then, you want to find two number $A$ and $B$ such that:

$$ \frac{y}{(y-1)(y+1)}=\frac{A}{y-1}+\frac{B}{y+1} $$

It is now easy to see that $A=B=\frac{1}{2}$ so that the first member equals the second.

Answer 2

To simplify things, put $x^2=t $.

then it is easy to see that

$$\frac {2t}{(t-1)(t+1)}=\frac{1}{(t-1)}+\frac {1}{ (t+1)} $$

Answer 3

You can just do the calculations backwards: $$\frac{x^2}{(x^2 + 1)(x^2-1)} = \frac{1}{2} \frac{2x^2}{(x^2 + 1)(x^2-1)} = \frac{1}{2}\frac{x^2 + 1 + x^2 - 1}{(x^2 + 1)(x^2-1)}.$$ And the rest should be obvious.

Answer 4

We want to write the left hand side in this form $$\frac{x^2}{(x^2-1)(x^2+1)} = \frac A {x^2-1} + \frac B {x^2+1}$$

Multiplying through by $(x^2-1)(x^2+1)$ gives us \begin{align}x^2&=\frac{A(x^2-1)(x^2+1)}{x^2-1}+\frac{B(x^2+1)(x^2-1)}{x^2+1}\\ x^2&=(x^2+1)A+(x^2-1)B\\ x^2&=Ax^2+A+Bx^2-B\\ x^2&=(A+B)x^2+(A-B)\end{align}

We can then equate coefficients to say that \begin{align}1&=A+B\tag{$x^2$ term}\\ 0&=A-B\tag{constant term}\end{align}

This means that $A=B=\frac 12$ so we now have:

\begin{align}\frac{x^2}{(x^2-1)(x^2+1)}&=\frac{\frac 12}{x^2-1}+\frac{\frac 12}{x^2+1}\\ &=\frac 12 \left(\frac1{x^2-1}+\frac1{x^2+1}\right)\end{align}

This technique is called partial fraction decomposition

Answer 5

To go from left to right, first assume

$$\frac{x^2}{(x^2-1)(x^2+1)}=\frac{A}{x^2-1}+\frac{B}{x^2+1}$$

A practical trick to get A is multiplying everything by $(x^2-1)$ and make it zero, so one gets

$$\frac{x^2}{(x^2-1)(x^2+1)}(x^2-1)=\frac{A}{x^2-1}(x^2-1)+\frac{B}{x^2+1}(x^2-1)$$

Now cancel those zeros in the numerator and denominator in the lhs and the first fraction of the rhs and make the second fraction of the rhs zero: $$\frac{x^2}{(x^2+1)}=A$$

This happens at $(x^2-1)=0$, equivalently $x^2=1$; substitute in the lhs and you are done with A.

$$\frac{1}{1+1}=A$$

The same procedure multiplying everything by $(x^2+1)$ and then using $x^2=-1$ yields B.

(Edit after the comment)

Multiply everything by $(x^2+1)$

$$\frac{x^2}{(x^2-1)(x^2+1)}(x^2+1)=\frac{A}{x^2-1}(x^2+1)+\frac{B}{x^2+1}(x^2+1)$$

The remaining fractions after cancelling are $$\frac{x^2}{(x^2-1)}=B$$

The condition is $x^2=-1$ and substituting in the lhs gives

$$\frac{-1}{-1-1}=\frac{1}{2}=B$$

This is the right real value for B although $x^2=-1$ corresponds to an imaginary $x$.