int {frac{{{e^x}(x - 1)}}{{{x^2}}};dx = } | vedclass

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Question

(A) We use the standard integral formula: $\int {e^x [f(x) + f'(x)]\,dx = e^x f(x) + c}$.Given the integral: $I = \int {\frac{{{e^x}(x - 1)}}{{{x^2}}}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We use the standard integral formula: $\int {e^x [f(x) + f'(x)]\,dx = e^x f(x) + c}$.Given the integral: $I = \int {\frac{{{e^x}(x - 1)}}{{{x^2}}}\;dx}$.Rewrite the integrand as: $I = \int {e^x \left( {\frac{x}{{{x^2}}} - \frac{1}{{{x^2}}}} \right)\,dx} = \int {e^x \left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)\,dx}$.Let $f(x) = \frac{1}{x}$. Then $f'(x) = -\frac{1}{{{x^2}}}$.Substituting this into the formula,we get: $I = e^x \left( \frac{1}{x} \right) + c = \frac{{{e^x}}}{x} + c$.

$\int {\frac{{{e^x}(x - 1)}}{{{x^2}}}\;dx = } $

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