Question

Resolve $\frac{x^4}{x^3+1}$ into partial fractions.

• A. $x+\frac{2}{(x+1)}-\frac{(x+1)}{3(x^2-x+1)}$
• B. $x+\frac{1}{(x+1)}-\frac{(x+1)}{3(x^2-x+1)}$
• C. $x-\frac{1}{3(x+1)}+\frac{(x+1)}{3(x^2-x+1)}$
• D. $x+\frac{1}{3(x+1)}-\frac{(x+1)}{3(x^2-x+1)}$

Answer 1

Answer: (D)

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

Given fraction is an improper fraction then by actual division
$x^3+1\downharpoonleft x^4\downharpoonright x$
i.e., $x^4+x$
$--$
_________
$-x$
$\therefore \frac{x^4}{x^3+1}=x-\frac{x}{x^3+1}$
$=x-\frac{x}{(x+1)(x^2-x+1)}$
Let $\frac{x^4}{x^3+1}=x-\{\frac{A}{x+1}+\frac{Bx+C}{x^2-x+1}\}$ ...(i)
or $x^4=x(x^3+1)-A(x^2-x+1)-(Bx+C)(x+1)$ ...(ii)
Putting $x+1=0$ or $x=-1$ in (ii) we obtain
$(-1{)}^4=0-A(1+1+1)-0$
$\therefore A=-\frac{1}{3}$
and putting $x^2-x+1=0$ or $x^2=x-1$ in (ii), we obtain
$(x-1{)}^2=0-0-B(x-1)-Bx-Cx-C$
$⟹x^2-2x+1=-2Bx-Cx+B-C$
Again Putting $x^2=x-1$
then $x-1-2x+1=-2Bx-Cx+B-C$
or $-x=-(2B+C)x+B-C$
On comparing
$2B+C=1$
and $B-C=0$
$\therefore B=C=\frac{1}{3}$
Substituting the values of A, B and C in (i), then
$\frac{x^4}{x^3+1}=x+\frac{1}{3(x+1)}-\frac{(x+1)}{3(x^2-x+1)}$
which are the required partial fractions.
Hence, option 'D' is correct.