Question

Resolve into partial fractions $\frac{x^2+1}{x(x^2-1)}$. Find the sum of coefficients of the individual terms.

Answer 1

Let $y=\frac{x^2+1}{x(x^2-1)}$

$x(x^2-1)=x(x+1)(x-1)$

Let $y=\frac{x^2+1}{x(x^2-1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}$

$\therefore x^2+1=A(x+1)(x-1)+Bx(x-1)+Cx(x+1)$

At $x=0,0^2+1=-A+0+0$

$\therefore A=-1$

At $x=1,1^2+1=0+0+2C$

$\therefore C=1$

At $x=-1,(-1{)}^2+1=0+B(-1\left)\right(-2)+0$

$\therefore B=1$

$\frac{x^2+1}{x(x^2-1)}=\frac{-1}{x}+\frac{1}{x+1}+\frac{1}{x-1}$

Sum of the coefficients are$A+B+C=1$