Question

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

Answer 1

$\frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}$
Let $\frac{2x+1}{x^2-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}$
$\Rightarrow 2x+1=A(x-1)+B(x+1)$ ........ (1)
Substituting $x=1$ and $-1$ in equation (1), we obtain
$A=\frac{1}{2}$ and $B=\frac{3}{2}$
$\therefore \frac{x^3+x+1}{x^2-1}=x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)}$
$\Rightarrow \int \frac{x^3+x+1}{x^2-1}dx=\int xdx+\frac{1}{2}\int \frac{1}{x+1}dx+\frac{3}{2}\int \frac{1}{(x-1)}dx$
$=\frac{x^2}{2}+\frac{1}{2}\log |x+1|+\frac{3}{2}\log |x-1|+C$