If (ex+2)(ex−1)(2ex−3)=−3ex−1+B2ex−3 , then B=

ex+2(ex−1)(2ex−3)=−3ex−1+B2ex−3 ex+2=−3(2ex−3)+B(ex−1) Put ex=32 then B=7

If (ex+2)(ex−1)(2ex−3)=−3ex−1+B2ex−3 , then B=

ex+2(ex−1)(2ex−3)=−3ex−1+B2ex−3 ex+2=−3(2ex−3)+B(ex−1) Put ex=32 then B=7

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

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

$e^{x} + 2 = - 3 (2 e^{x} - 3) + B (e^{x} - 1)$

Put $e^{x} = \frac{3}{2}$ then $B = 7$

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If (ex+2)(ex−1)(2ex−3)=−3ex−1+B2ex−3 , then B=

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