Integrate the rational function:
$\frac{x^{3}+x+1}{x^{2}-1}$

(N/A) The given integrand is an improper fraction because the degree of the numerator is greater than the degree of the denominator.
First,divide $(x^{3}+x+1)$ by $(x^{2}-1)$:
$\frac{x^{3}+x+1}{x^{2}-1} = x + \frac{2x+1}{x^{2}-1}$
Now,decompose $\frac{2x+1}{x^{2}-1}$ into partial fractions:
$\frac{2x+1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$
$2x+1 = A(x+1) + B(x-1)$
Setting $x=1$,we get $3 = 2A \Rightarrow A = \frac{3}{2}$.
Setting $x=-1$,we get $-1 = -2B \Rightarrow B = \frac{1}{2}$.
Thus,$\frac{x^{3}+x+1}{x^{2}-1} = x + \frac{3}{2(x-1)} + \frac{1}{2(x+1)}$.
Integrating both sides:
$\int \left( x + \frac{3}{2(x-1)} + \frac{1}{2(x+1)} \right) dx = \frac{x^{2}}{2} + \frac{3}{2} \ln|x-1| + \frac{1}{2} \ln|x+1| + C$.