Question

If $\frac{x^2+x+1}{(x+1{)}^2}=A+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}$ then $A+B+C$ is equal to

• A. 3
• B. 2
• C. 0
• D. 1

Answer 1

The correct option is D 1

Given $\frac{x^2+x+1}{(x+1{)}^2}=A+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}$

we can see that

$Numerator\left(x\right)+x=Denominator\left(x\right)$

$\Rightarrow x^2+x+1+x=(x+1{)}^2$

$\Rightarrow x^2+2x+1=x^2+2x+1$

So, $\frac{x^2+x+1}{(x+1{)}^2}=\frac{(x+1{)}^2-x}{(x+1{)}^2}$

$\frac{x^2+x+1}{(x+1{)}^2}=1-\frac{x}{(x+1{)}^2}--\left(1\right)$

Substitution technique $x+1=k$ in eqn (1)

$=1-\left(\frac{k-1}{k^2}\right)$

$=1-\frac{1}{k}+\frac{1}{k^2}$

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

by comparing with given we get

$A=1;B=-1;c=1$

$\therefore A+B+C=1$