Question

Resolve $\frac{42-19x}{(x^2+1)(x-4)}$ into partial fractions.

Answer 1

Assume $\frac{42-19x}{(x^2+1)(x-4)}=\frac{Ax+B}{x^2+1}+\frac{C}{x-4}$
$\therefore$ $42-19x=(Ax+B)(x-4)+C(x^2+1)$
Let $x=4$, then $C=-2$
Equating coefficients of $x^2$, $0=A+C$ and $A=2$,
Equating the absolute terms, $43=-4B+C$ and $B=-11$,
$\therefore \frac{42-19x}{(x^2+1)(x-4)}=\frac{2x-11}{x^2+1}-\frac{2}{x-4}$