Question

Resolve $\frac{x}{x^4+x^2+1}$ into partial fractions.

• A. $-\frac{1}{2(x^2+x+1)}+\frac{1}{2(x^2-x+1)}$
• B. $\frac{1}{2(x^2+x+1)}-\frac{1}{2(x^2-x+1)}$
• C. $-\frac{1}{(x^2+x+1)}+\frac{1}{(x^2-x+1)}$
• D. $-\frac{1}{4(x^2+x+1)}+\frac{1}{4(x^2-x+1)}$

Answer 1

Answer: (A)

$-\frac{1}{2(x^2+x+1)}+\frac{1}{2(x^2-x+1)}$

We have
$\frac{x}{x^4+x^2+1}=\frac{x}{(x^2+x+1)(x^2-x+1)}$ (Remember)
Let $\frac{x}{(x^2+x+1)(x^2-x+1)}=\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{x^2-x+1}$ ...(i)

$⟹x=(Ax+B)(x^2-x+1)+(Cx+D)(x^2+x+1)$ ...(ii)
Putting $x^2-x+1=0$
or $x^2=x-1$ in (ii), we obtain
$x=0+(Cx+D)(x-1+x+1)$
$⟹x=(Cx+D)2x$
$⟹\frac{1}{2}=Cx+D$
On comparing, $C=0$ and $D=\frac{1}{2}$
Again putting $x^2+x+1=0$
or $x^2=-x-1$ in (ii), we obtain
$x=(Ax+B)(-x-1-x+1)+0$
$⟹x=(Ax+B)(-2x)$
$⟹-\frac{1}{2}=Ax+B$
On Comparing, $A=0,B=-\frac{1}{2}$
Substituting the value of A, B, C, D in (i), we have
$\frac{x}{(x^2+x+1)(x^2-x+1)}=\frac{0-\frac{1}{2}}{x^2+x+1}+\frac{0+\frac{1}{2}}{x^2-x+1}$
or $\frac{x}{x^4+x^2+1}=-\frac{1}{2(x^2+x+1)}+\frac{1}{2(x^2-x+1)}$
which are the required partial fractions.
Hence, option 'A' is correct.