$$\int \frac{x^2+1}{(x)(x+1)(x-1)}$$
$\frac{x^2+1}{(x)(x+1)(x-1)} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-1} $
Through substitution, I found that
$A= -1 $ $B= 1$ $C= 1 $
And these are correct .
I'm just curious , so I tried the comparing coefficient method and I got stuck -
$x^2 + 1 = ax^2 - a + bx^2 - bx + cx^2 + cx $
Equation 1 - $A+B+C = 0$
Equation 2 - $ A = -1 $
Equation 3 - $ C - B = 0 $
From 3 - $ C = B $
Sub C=B into 1
$ -1 + B + B = 0 $ $ 2 B = 1 $ $B = 0.5$
Where did I go wrong ? Because B should be 1 .
By comparing coefficient of $x^2$, $$A+B+C=1$$
rather than
$$A+B+C=0$$