MCQ
$\int_{}^{} {\left( {1 + \frac{1}{{{x^2}}}} \right)\;{e^{\left( {x - \frac{1}{x}} \right)}}} \;dx$ =
- ✓${e^{x - \frac{1}{x}}} + c$
- B${e^{x + \frac{1}{x}}} + c$
- C${e^{{x^2} - \frac{1}{x}}} + c$
- D${e^{{x^2} + \frac{1}{{{x^2}}}}} + c$
Put $x - \frac{1}{x} = t \Rightarrow \left( {1 + \frac{1}{{{x^2}}}} \right)\,dx = dt$
$\therefore \,\,\,I = \int_{}^{} {{e^t}dt = {e^t} + c = {e^{x - \frac{1}{x}}} + c} $.
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