The value of ∫122e|x−1x|dx is equal to [Note: e denotes napier’s constant]

Let I=∫122e|x−1x|dx … (1) Put x=1y and adding, we get 2I=∫122e|x−1x|(1+1x2)dx =∫121e−(x−1x)(1+1x2)dx+∫12e(x−1x)(1+1x2)dx Put (x−1x)=t ⇒2I=2ee−2⇒I=(ee−1)

The value of ∫122e|x−1x|dx is equal to [Note: e denotes napier’s constant]

Let I=∫122e|x−1x|dx … (1) Put x=1y and adding, we get 2I=∫122e|x−1x|(1+1x2)dx =∫121e−(x−1x)(1+1x2)dx+∫12e(x−1x)(1+1x2)dx Put (x−1x)=t ⇒2I=2ee−2⇒I=(ee−1)

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The value of $\int_{\frac{1}{2}}^{2} e^{| x - \frac{1}{x} |} d x$ is equal to
[Note: e denotes napier’s constant]

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$e \sqrt{e} - 1$

$\sqrt{e} + 1$

$\sqrt{e} - 1$

$e \sqrt{e} + 1$

answer is B.

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Detailed Solution

Let    $I = \int_{\frac{1}{2}}^{2} e^{| x - \frac{1}{x} |} d x$ … (1)
Put $x = \frac{1}{y}$ and adding, we get
   $2 I = \int_{\frac{1}{2}}^{2} e^{| x - \frac{1}{x} |} (1 + \frac{1}{x^{2}}) d x = \int_{\frac{1}{2}}^{1} e^{- (x - \frac{1}{x})} (1 + \frac{1}{x^{2}}) d x + \int_{1}^{2} e^{(x - \frac{1}{x})} (1 + \frac{1}{x^{2}}) d x$
Put    $(x - \frac{1}{x}) = t$ $\Rightarrow 2 I = 2 e \sqrt{e} - 2 \Rightarrow I = (e \sqrt{e} - 1)$