Mathematics

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


ANSWER

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


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
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