Question

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

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

Answer 1

The correct option is A $\frac{-1}{4(1+x)}+\frac{(x-1)}{4(1+x^2)}+\frac{(x+1)}{2(1+x^2{)}^2}$

Let $\frac{x}{(1+x)(1+x^2{)}^2}=\frac{A}{1+x}+\frac{Bx+C}{(1+x^2)}+\frac{(Dx+E)}{(1+x^2{)}^2}$ ...(i)
$⟹x=A(1+x^2{)}^2+(Bx+C\left)\right(1+x\left)\right(1+x^2)+(Dx+E\left)\right(1+x)$ ...(ii)
Putting $1+x=0$ or $x=-1$ in (ii), we obtain
$-1=A[1+(-1{)}^2{]}^2+0+0$
$\therefore A=-\frac{1}{4}$
Putting $1+x^2=0$ or $x^2=-1$ in (ii), we obtain
$x=0+0+Dx+D(-1)+E+Ex$
$⟹x=(D+E)x+(E-D)$
Equating the coefficient of $x$ and constant term, we get
$D+E=1$
and $E-D=0$
$\therefore D=E=\frac{1}{2}$
Comparing the constant terms in (ii), we obtain
$0=A+C+E$ (For comparing constant terms putting $x=0$)
or $0=-\frac{1}{4}+C+\frac{1}{2}$
$⟹0=C+\frac{1}{4}$
$\therefore C=-\frac{1}{4}$
and comparing the coefficient of $x^4$ in (ii), we obtain
$0=A+B$
$\therefore B=-A$
$B=\frac{1}{4}$
Substituting the value of A, B, C, D, and E in (1), then
$\frac{x}{(1+x)(1+x^2{)}^2}=\frac{-1}{4(1+x)}+\frac{(x-1)}{4(1+x^2)}+\frac{(x+1)}{2(1+x^2{)}^2}$
which are the required partial fractions.
Hence, option 'A' is correct.