Answer
$\dfrac{x^5+2}{x^2-1}=x^3+x+\dfrac{3}{2(x-1)}-\dfrac{1}{2(x+1)}$
Work Step by Step
We are given the fraction:
$\dfrac{x^5+2}{x^2-1}$
Because the degree of the numerator is greater than the degree of the denominator, we divide the numerator by the denominator:
$\dfrac{x^5+2}{x^2-1}=\dfrac{x^5-x^3+x^3-x+x+2}{x^2-1}=\dfrac{x^3(x^2-1)+x(x^2-1)+x+2}{x^2-1}$
$=\dfrac{(x^2-1)(x^3+x)+x+2}{x^2-1}=x^3+x+\dfrac{x+2}{x^2-1}$
We will write the partial fraction decomposition for the fraction $\dfrac{x+2}{x^2-1}$. Factor the denominator and write the partial fraction decomposition:
$\dfrac{x+2}{x^2-1}=\dfrac{x+2}{(x-1)(x+1)}=\dfrac{A}{x-1}+\dfrac{B}{x+1}$
Identify the coefficients:
$x+2=A(x+1)+B(x-1)$
$x+2=Ax+A+Bx-B$
$x+2=(A+B)x+(A-B)$
$\begin{cases}
A+B=1\\
A-B=2
\end{cases}$
$A+B+A-B=1+2$
$2A=3$
$A=\dfrac{3}{2}$
$A+B=1$
$\dfrac{3}{2}+B=1$\\
$B=1-\dfrac{3}{2}$
$B=-\dfrac{1}{2}$
The partial fraction decomposition is:
$\dfrac{x+2}{(x-1)(x+1)}=\dfrac{\dfrac{3}{2}}{x-1}-\dfrac{\dfrac{1}{2}}{x+1}$
$\dfrac{x+2}{(x-1)(x+1)}=\dfrac{3}{2(x-1)}-\dfrac{1}{2(x+1)}$
The final result is:
$\dfrac{x^5+2}{x^2-1}=x^3+x+\dfrac{3}{2(x-1)}-\dfrac{1}{2(x+1)}$
