Question

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

• A. $-\frac{2}{x^2}+\frac{1}{(x^2+1{)}^2}$
• B. $-\frac{1}{x^2}+\frac{3}{(x^2+1{)}^2}$
• C. $\frac{1}{x^2}-\frac{5}{(x^2+1{)}^2}$
• D. $\frac{1}{x^2}-\frac{3}{(x^2+1{)}^2}$

Answer 1

The correct option is D $\frac{1}{x^2}-\frac{3}{(x^2+1{)}^2}$

The given fraction has $x^2$ everywhere then put $x^2=t$ for the sake of partial fractions.
$\therefore \frac{x^4-x^2+1}{x^2(x^2+1{)}^2}=\frac{t^2-t+1}{t(t+1{)}^2}$
Now let $\frac{t^2-t+1}{t(t+1{)}^2}=\frac{A}{t}+\frac{B}{t+1}=\frac{C}{(t+1{)}^2}$ ...(i)
$⟹t^2-t+1=A(t+1{)}^2+Bt(t+1)+Ct$ ...(ii)
Substituting $t+1=0$ or $t=-1$ in (ii), we ontain
$(-1{)}^2-(-1)+1=0+0+C(-1)$
$⟹C=-3$
Substituting $t=0$ in (ii) we obtain
$0-0+1=A(0+1{)}^2+0+0$
$⟹A=1$
Now equating the coefficient of $t^2$ on both sides of (ii) then
$1=A+B$
$⟹1=1+B$
$\therefore B=0$
Substituting the values of A, B and C in (i) we get
$\frac{t^2-t+1}{t(t+1{)}^2}=\frac{1}{t}-\frac{3}{(t+1{)}^2}$
In last substitution $t=x^2$
then $\frac{x^4-x^2+1}{x^2(x^2+1{)}^2}=\frac{1}{x^2}-\frac{3}{(x^2+1{)}^2}$
which are the required partial fractions.
Hence, option 'D' is correct.