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STD 12 - 7.1 inderfinite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.1 inderfinite integral4 Marks
MCQ
The value of $\int_{}^{} {\left( {1 + \frac{1}{{{x^2}}}} \right)\;{e^{\left( {x - \frac{1}{x}} \right)}}} \;dx$ equals
  • ✓
    ${e^{x - \frac{1}{x}}} + c$
  • B
    ${e^{x + \frac{1}{x}}} + c$
  • C
    ${e^{{x^2} - \frac{1}{x}}} + c$
  • D
    ${e^{{x^2} + \frac{1}{{{x^2}}}}} + c$
✓

Answer

Correct option: A.
${e^{x - \frac{1}{x}}} + c$
a
(a) $I = \int_{}^{} {\left( {1 + \frac{1}{{{x^2}}}} \right){\rm{ }}{e^{x - \frac{1}{x}}}dx} $

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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