Question

If $\frac{(x-1{)}^2}{x^3+x}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$, then $A=...,B=...,C=...$

• A. $A=1,B=0,C=-2$
• B. $A=1,B=0,C=2$
• C. $A=0,B=1,C=-2$
• D. $A=0,B=1,C=2$

Answer 1

The correct option is A $A=1,B=0,C=-2$

$\frac{(x-1{)}^2}{x^3+x}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$

$\Rightarrow {(x-1)}^2=A(x^2+1)+(Bx+C)x$

$\Rightarrow (x^2-2x+1)=A(x^2+1)+B{x}^2+Cx$

$\Rightarrow (x^2-2x+1)=(A+B)x^2+Cx+A$

On comparing coefficients

$A+B=1,C=-2,A=1\Rightarrow A=1,B=0,C=-2$

Hence, option 'A' is correct.