The correct option is A e^x+1/x+c To find #160; ∫ (1 1x^2)e^(x+1x) dx Substitute x+1x=t (1 1x^2)· dx = dt ∫ e ^t· dt #160;· #16 ...

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Property 2

∫1-1/x2ex+1/x...

Question

$\int (1 - \frac{1}{x^{2}}) e^{x + \frac{1}{x}} d x$

e^{x + \frac{1}{x}} + c

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1 - \frac{1}{x^{2}} + c

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- e^{x + \frac{1}{x}} + c

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1 + \frac{1}{x^{2}} + c

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Solution

The correct option is A $e^{x + \frac{1}{x}} + c$
To find $\int (1 - \frac{1}{x^{2}}) e^{(x + \frac{1}{x})} d x$
Substitute $x + \frac{1}{x} = t$
$(1 - \frac{1}{x^{2}}) \cdot d x = d t$
$\int e^{t} \cdot d t \cdot e^{t} + c$
$= e^{(x + \frac{1}{x})} + c$
$\int (1 - \frac{1}{x^{2}}) e^{(x + \frac{1}{x})} \cdot = e^{(x + \frac{1}{x})} + c$

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