Question

Resolve into partial fractions :$\frac{7+x}{(1+x)(1+x^2)}$

Answer 1

Let, $\frac{7+x}{(1+x)(1+x^2)}=\frac{A}{1+x}+\frac{Bx+C}{1+x^2}$

$⟹\frac{7+x}{(1+x)(1+x^2)}=\frac{A(1+x^2)+(Bx+C)(1+x)}{(1+x)(1+x^2)}$

$⟹7+x=A+A{x}^2+Bx+B{x}^2+C+Cx$

$⟹7+x=(A+B)x^2+(B+C)x+(A+C)$

i.e., $A+B=0$...........(i)

$B+C=1$

and $A+C=7$

$⟹A-B=6$........(ii)

By (i) & (ii) we get, $A=3,B=-3$ and $C=4$

$\therefore \frac{7+x}{(1+x)(1+x^2)}=\frac{3}{1+x}+\frac{-3x+4}{1+x^2}$