Answer to Question #342459 in Calculus for Dkay

Question

Decompose (x^2+x+1)/((x+3)(x^2-x+1))

Answer 1

$\frac{x^2+x+1}{(x+3)(x^2-x+1)}=\frac{A}{x+3}+\frac{Bx+C}{x^2-x+1}$

$x^2+x+1=A(x^2-x+1)+(Bx+C)(x+3)$

Let's find A, B and C by equating corresponding coefficients:

$1=A+B$ (1)

$1=-A+3B+C$ (2)

$1=A+3C$ (3)

Substracting (3) from (1) we obtain:

$B=3C$ (4)

Substitution (4) into (2) gives

$1=-A+10C$ (5)

Adding (3) and (5) we obtain

$C=\frac{2}{13}$

$B=\frac{6}{13}$

$A=\frac{7}{13}$

Answer:

$\frac{x^2+x+1}{(x+3)(x^2-x+1)}=\frac{7}{13}\frac{1}{x+3}+\frac{2}{13}\frac{3x+1}{x^2-x+1}$