Mathematics

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


ANSWER

$$\displaystyle \frac{-1}{4(1+x)}+\frac{(x-1)}{4(1+x^2)}+\frac{(x+1)}{2(1+x^2)^2}$$


SOLUTION
Let             $$\displaystyle \frac{x}{(1+x)(1+x^2)^2}=\frac{A}{1+x}+\frac{Bx+C}{(1+x^2)}+\frac{(Dx+E)}{(1+x^2)^2}$$                      ...(i)
$$\Longrightarrow                  x=A(1+x^2)^2+(Bx+C)(1+x)(1+x^2)+(Dx+E)(1+x)$$                ...(ii)
Putting $$1+x=0$$ or $$x=-1$$ in (ii), we obtain
          $$-1=A[1+(-1)^2]^2+0+0$$
$$\therefore                                         A=-\dfrac{1}{4}$$
Putting $$1+x^2=0$$ or $$x^2= -1$$ in (ii), we obtain
               $$x=0+0+Dx+D(-1)+E+Ex$$
$$\Longrightarrow               x=(D+E)x+(E-D)$$
Equating the coefficient of $$x$$ and constant term, we get
                           $$D+E=1$$
and    $$E-D=0$$
$$\therefore                                        D=E=\dfrac{1}{2}$$
Comparing the constant terms in (ii), we obtain
           $$0=A+C+E$$   (For comparing constant terms putting $$x=0$$)
or                 $$0=-\dfrac{1}{4}+C+\dfrac{1}{2}$$
$$\Longrightarrow                     0=C+\dfrac{1}{4}$$
$$\therefore                             C=-\dfrac{1}{4}$$
and comparing the coefficient of $$x^4$$ in (ii), we obtain 
               $$0=A+B$$
$$\therefore                    B= -A$$
               $$B=\dfrac{1}{4}$$
Substituting the value of A, B, C, D, and E in (1), then
          $$\displaystyle \frac{x}{(1+x)(1+x^2)^2}= \frac{-1}{4(1+x)}+\frac{(x-1)}{4(1+x^2)}+\frac{(x+1)}{2(1+x^2)^2}$$
which are the required partial fractions.
Hence, option 'A' is correct.
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Single Correct Medium Published on 17th 09, 2020
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