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
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
$4$
$5$
$6$
$8$
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}
$
