The value of int {left( {1 + frac{1}{{{x^2}}} | vedclass

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Question

(A) Let $I = \int {\left( {1 + \frac{1}{{{x^2}}}} \right){e^{x - \frac{1}{x}}}} \,dx$. Substitute $t = x - \frac{1}{x}$. Differentiating both sides wi

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

Vedclass · Answer

(A) Let $I = \int {\left( {1 + \frac{1}{{{x^2}}}} \right){e^{x - \frac{1}{x}}}} \,dx$. Substitute $t = x - \frac{1}{x}$. Differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = 1 - ( - \frac{1}{{{x^2}}}) = 1 + \frac{1}{{{x^2}}}$. Therefore,$(1 + \frac{1}{{{x^2}}})\,dx = dt$. Substituting these into the integral,we get $I = \int {{e^t}} \,dt$. The integral of ${e^t}$ is ${e^t} + c$. Substituting back $t = x - \frac{1}{x}$,we get $I = {e^{x - \frac{1}{x}}} + c$.

The value of $\int {\left( {1 + \frac{1}{{{x^2}}}} \right){e^{\left( {x - \frac{1}{x}} \right)}}} \,dx$ equals

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