Question

If $\frac{(e^x+2)}{(e^x-1)(2e^x-3)}=\frac{-3}{e^x-1}+\frac{B}{2e^x-3}$ then B=

• A. $1$
• B. $3$
• C. $5$
• D. $7$

Answer 1

The correct option is D $7$

Given, $\frac{(e^x+2)}{(e^x-1)(2e^x-3)}=\frac{3}{e^x-1}+\frac{B}{2e^x-3}$ ....(1)
Consider, $\frac{(e^x+2)}{(e^x-1)(2e^x-3)}$
Let $e^x=t$ and resolving into partial fractions
$\frac{(t+2)}{(t-1)(2t-3)}=\frac{A}{t-1}+\frac{B}{2t-3}$ ....(2)
$t+2=A(2t-3)+B(t-1)$
$\Rightarrow 2A+B=1,-3A-B=2$
$\Rightarrow A=-3,B=7$
Put this value in (2)
$\frac{(t+2)}{(t-1)(2t-3)}=\frac{-3}{t-1}+\frac{7}{2t-3}$
$\frac{(e^x+2)}{(e^x-1)(2e^x-3)}=\frac{-3}{e^x-1}+\frac{7}{2e^x-3}$