If int f(x) , dx = x e^{-log |x|} + f(x),then | vedclass

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(C) Given the equation $\int f(x) \, dx = x e^{-\log |x|} + f(x)$.We know that $e^{-\log |x|} = e^{\log |x|^{-1}} = |x|^{-1} = \frac{1}{|x|}$.Substitu

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$c e^x$

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(C) Given the equation $\int f(x) \, dx = x e^{-\log |x|} + f(x)$.We know that $e^{-\log |x|} = e^{\log |x|^{-1}} = |x|^{-1} = \frac{1}{|x|}$.Substituting this into the equation,we get $\int f(x) \, dx = x \cdot \frac{1}{|x|} + f(x)$.Assuming $x > 0$,$|x| = x$,so $\int f(x) \, dx = 1 + f(x)$.Differentiating both sides with respect to $x$,we get $f(x) = 0 + f'(x)$,which implies $f'(x) = f(x)$.This is a first-order linear differential equation $\frac{df}{dx} = f(x)$,which can be written as $\frac{df}{f} = dx$.Integrating both sides,we get $\log |f(x)| = x + C$,which implies $f(x) = e^{x+C} = c e^x$.

If $\int f(x) \, dx = x e^{-\log |x|} + f(x)$,then $f(x)$ is

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