WAEC Further Mathematics 2013 Paper

Concept:Use partial fraction decomposition because the denominator contains a linear factor $(1-x)$ and an irreducible quadratic factor $(x^{2}+1)$.Explanation:Write the expression in the standard partial fraction form:$$\frac{x^{2}+x+4}{(1-x)(x^{2}+1)} = \frac{A}{1-x} + \frac{Bx+C}{x^{2}+1}.$$Multiply both sides by the denominator $(1-x)(x^{2}+1)$:$$x^{2}+x+4 = A(x^{2}+1) + (Bx+C)(1-x).$$Expand the right-hand side:$$x^{2}+x+4 = Ax^{2}+A + Bx - Bx^{2} + C - Cx.$$Group like terms:$$x^{2}+x+4 = (A-B)x^{2} + (B-C)x + (A+C).$$Compare coefficients of $x^{2}$, $x$, and the constant term:$$A - B = 1, \quad B - C = 1, \quad A + C = 4.$$From $A - B = 1$, get $A = B + 1$.From $B - C = 1$, get $C = B - 1$.Substitute into $A + C = 4$:$$(B+1) + (B-1) = 4.$$$$2B = 4 \Rightarrow B = 2.$$Therefore, $A = 3$ and $C = 1$.Thus, the partial fraction decomposition is:$$\frac{x^{2}+x+4}{(1-x)(x^{2}+1)} = \frac{3}{1-x} + \frac{2x+1}{x^{2}+1}.$$Answer:Option B: $\frac{3}{1-x} + \frac{2x+1}{x^{2}+1}$.