int e^{x} frac{(x-1)}{x^{2}} d x is equal to | vedclass

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(C) We know that $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c$.The given integral is $\int e^{x} \left(\frac{x-1}{x^{2}}\right) dx$.This can be rewr

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

Vedclass · Answer

(C) We know that $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c$.The given integral is $\int e^{x} \left(\frac{x-1}{x^{2}}\right) dx$.This can be rewritten as $\int e^{x} \left(\frac{1}{x} - \frac{1}{x^{2}}\right) dx$.Here,let $f(x) = \frac{1}{x}$,then $f'(x) = -\frac{1}{x^{2}}$.Applying the formula,we get $\int e^{x} \left(\frac{1}{x} - \frac{1}{x^{2}}\right) dx = e^{x} \left(\frac{1}{x}\right) + c = \frac{e^{x}}{x} + c$.

$\int e^{x} \frac{(x-1)}{x^{2}} d x$ is equal to

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