∫ex−e[x]−x−2⋅e3x2dx∀x∈(0,1) is equal to [where [x] is greatest integer of x

Let I=∫ex−e[x]−x−2e3x2dx,x∈(0,1)I=∫ex−e−x−2e3x2dxI=∫ex2−2ex−1e3x2dx Put ex=tI=∫t2−2t−1 dtI=∫(t−1)2−(2)2dt=(t−1)2(t−1)2−(2)2−(2)22log⁡(t−1)+(t−1)2−(2)2+C=ex−12ex−12−2−log⁡ex−1+ex−12−2+C=ex−12e2x−2ex−1−log⁡ex−1+e2x−2ex−1+C

∫ex−e[x]−x−2⋅e3x2dx∀x∈(0,1) is equal to [where [x] is greatest integer of x

Let I=∫ex−e[x]−x−2e3x2dx,x∈(0,1)I=∫ex−e−x−2e3x2dxI=∫ex2−2ex−1e3x2dx Put ex=tI=∫t2−2t−1 dtI=∫(t−1)2−(2)2dt=(t−1)2(t−1)2−(2)2−(2)22log⁡(t−1)+(t−1)2−(2)2+C=ex−12ex−12−2−log⁡ex−1+ex−12−2+C=ex−12e2x−2ex−1−log⁡ex−1+e2x−2ex−1+C

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

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

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

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

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

answer is D.

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Detailed Solution

$Let I = \int \sqrt{e^{x} - e^{[x] - x} - 2} e^{\frac{3 x}{2}} d x, x \in (0, 1)$

$\begin{aligned} I = \int \sqrt{e^{x} - e^{- x} - 2} e^{\frac{3 x}{2}} d x \\ I = \int \sqrt{{(e^{x})}^{2} - 2 e^{x} - 1} e^{\frac{3 x}{2}} d x \end{aligned}$

$Put e^{x} = t$

$I = \int \sqrt{t^{2} - 2 t - 1} d t$

$\begin{array}{l} I = \int \sqrt{(t - 1)^{2} - (\sqrt{2})^{2}} d t \\ = \frac{(t - 1)}{2} \sqrt{(t - 1)^{2} - (\sqrt{2})^{2}} - \frac{(\sqrt{2})^{2}}{2} \log |(t - 1) + \sqrt{(t - 1)^{2} - (\sqrt{2})^{2}}| + C \\ = \frac{e^{x} - 1}{2} \sqrt{{(e^{x} - 1)}^{2} - 2} - \log |e^{x} - 1 + \sqrt{{(e^{x} - 1)}^{2} - 2}| + C \\ = \frac{e^{x} - 1}{2} \sqrt{e^{2 x} - 2 e^{x} - 1} - \log |e^{x} - 1 + \sqrt{e^{2 x} - 2 e^{x} - 1}| + C \end{array}$