Question

Integrate the rational function $\frac{3x+5}{x^3-x^2-x+1}$

Answer 1

$\frac{3x+5}{x^3-x^2-x+1}=\frac{3x+5}{(x-1{)}^2(x+1)}$
Let $\frac{3x+5}{(x-1{)}^2(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1{)}^2}+\frac{C}{(x+1)}$
$\Rightarrow 3x+5=A(x-1)(x+1)+B(x+1)+C(x-1{)}^2$
$\Rightarrow 3x+5=A(x^2-1)+B(x+1)+C(x^2+1-2x)$ ......... (1)
Substituting $x=1$ in equation (1), we obtain
$B=4$
Equating the coefficients of $x^2$ and $x$, we obtain
$A+C=0$
$B-2C=3$
On solving, we obtain
$A=-\frac{1}{2}$ and $C=\frac{1}{2}$
$\therefore \frac{3x+5}{(x-1{)}^2(x+1)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1{)}^2}+\frac{1}{2(x+1)}$
$\Rightarrow \int \frac{3x+5}{(x-1{)}^2(x+1)}dx=-\frac{1}{2}\int \frac{1}{x-1}dx+4\int \frac{1}{(x-1{)}^2}dx+\frac{1}{2}\int \frac{1}{(x+1)}dx$
$=-\frac{1}{2}\log |x-1|+4\left(\frac{-1}{x-1}\right)+\frac{1}{2}\log |x+1|+C$
$=\frac{1}{2}\log \left|\frac{x+1}{x-1}\right|-\frac{4}{(x-1)}+C$