WAEC Further Mathematics 2013 Paper

$\frac{x^{2}+x+4}{(1-x)(x^{2}+1)} = \frac{A}{1-x} + \frac{Bx+C}{x^{2}+1}$

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

$\Rightarrow x^{2}+x+4 = A(x^{2}+1)+(Bx+C)(1-x)$

$x^{2}+x+4 = Ax^{2}+A+Bx-Bx^{2}-Cx+C$

$\Rightarrow (A-B)x^{2}=x^{2};\ A-B=1\ \dots\ (i)$

$(B-C)x=x;\ B-C=1\ \dots\ (ii)$

$A+C=4\ \dots\ (iii)$

Solving the above simultaneous equations by any of the known methods, we get

$A=3,\ B=2,\ C=1$

$\therefore \frac{x^{2}+x+4}{(1-x)(x^{2}+1)} = \frac{3}{1-x} + \frac{2x+1}{x^{2}+1}$