$\int \sqrt{e^{x} - e^{[x] - x} - 2} \cdot e^{\frac{3 x}{2}} d x \forall x \in (0, 1) is equal to [where [x] is greatest integer of x]$

Difficult

A

$\frac{e^{x} - 1}{2} \sqrt{e^{2 x} - 2 e^{x} + 1} + \frac{1}{2} \log |e^{x} - 1 + \sqrt{e^{2 x} - 2 e^{x} + 1}| + C$
B

$\frac{e^{x} - 1}{2} \sqrt{e^{2 x} - 2 e^{x} - 1} + \frac{1}{2} \log |e^{x} - 1 + \sqrt{e^{2 x} - 2 e^{x} - 1}| + C$
C

$\frac{e^{x} - 1}{2} \sqrt{e^{2 x} - 2 e^{x} - 1} + \log |e^{x} - 1 + \sqrt{e^{2 x} - 2 e^{x} - 1}| + C$
D

$\frac{e^{x} - 1}{2} \sqrt{e^{2 x} - 2 e^{x} - 1} - \log |e^{x} - 1 + \sqrt{e^{2 x} - 2 e^{x} - 1}| + C$

Solution

$Let I = \int \sqrt{e^{x} - e^{[x] - x} - 2} e^{\frac{3 x}{2}} d x, x \in (0, 1)$
$\begin{aligned} I = \int \sqrt{e^{x} - e^{- x} - 2} e^{\frac{3 x}{2}} d x \\ I = \int \sqrt{{(e^{x})}^{2} - 2 e^{x} - 1} e^{\frac{3 x}{2}} d x \end{aligned}$
$Put e^{x} = t$
$I = \int \sqrt{t^{2} - 2 t - 1} d t$
$\begin{array}{l} I = \int \sqrt{(t - 1)^{2} - (\sqrt{2})^{2}} d t \\ = \frac{(t - 1)}{2} \sqrt{(t - 1)^{2} - (\sqrt{2})^{2}} - \frac{(\sqrt{2})^{2}}{2} \log |(t - 1) + \sqrt{(t - 1)^{2} - (\sqrt{2})^{2}}| + C \\ = \frac{e^{x} - 1}{2} \sqrt{{(e^{x} - 1)}^{2} - 2} - \log |e^{x} - 1 + \sqrt{{(e^{x} - 1)}^{2} - 2}| + C \\ = \frac{e^{x} - 1}{2} \sqrt{e^{2 x} - 2 e^{x} - 1} - \log |e^{x} - 1 + \sqrt{e^{2 x} - 2 e^{x} - 1}| + C \end{array}$

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