Explanation solution partial-fraction of $\frac{x^2 + 2}{x^2 - 1}$

Question

The partial fraction of $\dfrac{x^2+2}{x^2-1}$ is $1 + \dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$.

I understand how you get $\dfrac{3}{2}\cdot(\dfrac{1}{x-1}-\dfrac{1}{x+1})$ but from where does the $1 +$ come?

Answer 1

Hint. you may write $$ \frac{x^2+2}{x^2-1}=\frac{(x^2-1)+3}{x^2-1}=\frac{x^2-1}{x^2-1}+\frac{3}{x^2-1}=\color{red}{1+}\frac{3}{x^2-1} $$

Answer 2

Since the polynomial in the numerator has the same degree as the polynomial in the denominator the remained in their polynomial division is a non-zero polynomial (the constant 1 in this case).

Answer 3

1 comes from long division of $x^2+2$ by $x^2-1.$ when you divide $x^2 + 2$ by $x^2 - 1$ you get the quotient $1$ and the remainder $3.$ that is $$x^2 + 2 = 1(x^2-1)+3 \mbox{ or }{x^2 + 2 \over x^2 -1} = 1 + {3 \over x^2 - 1}$$

this is very similar to how you do division in integers. for example when you divide $123$ by $7,$ the quotient to be 17(how many times 7 goes into 123) and remainder(what is left over) to be 4. we can write this in two ways: $$123 = 7*17 + 4 \mbox{ or } {123 \over 7} = 17 + {4 \over 7}$$