If $\frac{(x+1)}{(2 x-1)(3 x+1)}=\frac{A}{(2 x-1)}+\frac{B}{(3 x+1)}$, then $16 A+9 B$ is equal to

• A

  $4$
• B

  $5$
• Correct answer
  C

  $6$
• D

  $8$

Solution

Given that, $$ \begin{aligned} & \frac{x+1}{(2 x-1)(3 x+1)}=\frac{A}{(2 x-1)}+\frac{B}{(3 x+1)} \\ \Rightarrow & (x+1)=A(3 x+1)+B(2 x-1) \\ \Rightarrow & (x+1)=x(3 A+2 B)+A-B \end{aligned} $$ On equating the coefficient of $x$ and constant on both sides, we get
On solving Eqs. (i) and (ii), we get $$ \begin{aligned} A & =\frac{3}{5}, B=-\frac{2}{5} \\ \therefore \quad 16 A+9 B & =16\left(\frac{3}{5}\right)+9\left(-\frac{2}{5}\right) \\ & =\frac{48}{5}-\frac{18}{5}=\frac{30}{5}=6 \end{aligned} $$

Hence, the correct answer is option: 'C'.