Mathematics

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

$\displaystyle \frac{1}{x^2}-\frac{3}{(x^2+1)^2}$

##### SOLUTION
The given fraction has $x^2$ everywhere then put $x^2=t$ for the sake of partial fractions.
$\therefore \displaystyle \frac{x^4-x^2+1}{x^2(x^2+1)^2}=\frac{t^2-t+1}{t(t+1)^2}$

Now let         $\displaystyle \frac{t^2-t+1}{t(t+1)^2} = \frac{A}{t}+\frac{B}{t+1}=\frac{C}{(t+1)^2}$                       ...(i)

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

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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