Question

Resolve $\frac{1}{x^4+1}$ into partial fractions.

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

Answer 1

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

We have $x^4+1=(x^2{)}^2+(1{)}^2+2x^2-2x^2$
$=(x^2+1{)}^2-(x\sqrt{2}{)}^2$
$=(x^2+x\sqrt{2}+1)(x^2-x\sqrt{2}+1)$
$\therefore \frac{1}{x^4+1}=\frac{1}{(x^2+x\sqrt{2}+1)(x^2-x\sqrt{2}+1)}$
Let $\frac{1}{(x^2+x\sqrt{2}+1)(x^2-x\sqrt{2}+1)}=\frac{Ax+B}{x^2+x\sqrt{2}+1}+\frac{Cx+D}{x^2+x\sqrt{2}+1}$ ...(i)
$⟹1=(Ax+B)(x^2-x\sqrt{2}+1)+(Cx+D)(x^2+x\sqrt{2}+1)$ ...(ii)
Putting $x^2-x\sqrt{2}+1=0$ or $x^2=x\sqrt{2}-1$ in (ii), we obtain
$1=0+(Cx+D)\left(2x\sqrt{2}\right)$
$1=2C\sqrt{2}{x}^2+2Dx\sqrt{2}$
$1=2C\sqrt{2}(x\sqrt{2}-1)+2D\sqrt{2}x$ ($\because x^2=x\sqrt{2}-1$)
$1=2.2Cx-2C\sqrt{2}+2\sqrt{2}Dx$
On comparing
$0=4c+2\sqrt{2}D$
$\therefore D=-\sqrt{2}C$
and $1=-2C\sqrt{2}$
$\therefore C=-\frac{1}{2\sqrt{2}},D=\frac{1}{2}$
and putting $x^2+x\sqrt{2}+1=0$ or $x^2=-x\sqrt{2}-1$ in (ii), we obtain
$1=(Ax+B)(-2\sqrt{2}x)+0$
$1=-2\sqrt{2}A{x}^2-2\sqrt{2}Bx$
$1=-2\sqrt{2}A(-x\sqrt{2}-1)-2\sqrt{2}Bx$
$1=4Ax+2\sqrt{2}A-2\sqrt{2}Bx$
on comparing $A=\frac{1}{2\sqrt{2}}$
and $4A-2\sqrt{2}B=0$
$\therefore B=\sqrt{2}A$
$\therefore B=\frac{1}{2}$
Substituting the values of A, B, C and D in (i), we have
$\frac{1}{x^4+1}=\frac{(x+\sqrt{2})}{2\sqrt{2}(x^2+x\sqrt{2}+1)}-\frac{(x-\sqrt{2})}{2\sqrt{2}(x^2-x\sqrt{2}+1)}$
which are the required partial fractions.
Hence, option 'D' is correct.