int {(1 + x - {x^{ - 1}}){e^{x + {x^{ - 1}}}} | vedclass

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Question

(D) Let $I = \int {(1 + x - {x^{ - 1}}){e^{x + {x^{ - 1}}}}\,dx}$.We can rewrite the integrand as:$I = \int {e^{x + {x^{ - 1}}}} (1 + x - \frac{1}{x})

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int {(1 + x - {x^{ - 1}}){e^{x + {x^{ - 1}}}}\,dx}$.We can rewrite the integrand as:$I = \int {e^{x + {x^{ - 1}}}} (1 + x - \frac{1}{x}) \,dx$.Observe that $\frac{d}{dx}(x{e^{x + {x^{ - 1}}}}) = 1 \cdot {e^{x + {x^{ - 1}}}} + x \cdot {e^{x + {x^{ - 1}}}} \cdot (1 - \frac{1}{x^2})$.$= {e^{x + {x^{ - 1}}}} [1 + x(1 - \frac{1}{x^2})] = {e^{x + {x^{ - 1}}}} [1 + x - \frac{1}{x}]$.Thus,$\int {e^{x + {x^{ - 1}}}} (1 + x - \frac{1}{x}) \,dx = x{e^{x + {x^{ - 1}}}} + c$.

$\int {(1 + x - {x^{ - 1}}){e^{x + {x^{ - 1}}}}\,dx} = $

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