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

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(D) We know that the integral of the form $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$. Here,let $f(x) = \frac{1}{x}$. Then,$f'(x) = -\frac{1}{x^2}$. S

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

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(D) We know that the integral of the form $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$. Here,let $f(x) = \frac{1}{x}$. Then,$f'(x) = -\frac{1}{x^2}$. Substituting these into the integral: $I = \int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx = \int e^x \left( f(x) + f'(x) \right) dx$. Using the formula,we get $I = e^x f(x) + c = \frac{e^x}{x} + c$.

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

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