Question

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

• A. 2
• B. -1
• C. 1
• D. 0

Answer 1

The correct option is D 0

$LHS=\frac{1}{(x+1)(x^2+2x+2)};RHS=\frac{A}{x+1}+\frac{Bx+C}{(x+1{)}^2+1}$

$\frac{A\left[\right(x+1{)}^2+1]+(Bx+C\left)\right(x+1)}{(x+1)\left[\right(x+1{)}^2+1]}$

comparing the coefficients of numerators in LHS and RHS in the eqn.

$1=A\left[\right(x+1{)}^2+1]+(Bx+C\left)\right(x+1)$

co-efficient of $x^2$ is zero

So. $A+B=0--\left(1\right)$

co-efficient of $x$

$2A+B+C=0--\left(2\right)$

Constant term $=1$

$2A+C=1--\left(3\right)$

Solving equation 1,2,3

$A+C=0$ from (1) & (2)

$2A+C=1$ (3)

So, $A=1;C=-1$

$B=-1$

$A+B=0$