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With f = 1/x and x > 0, e^x (f + f') integrates to

Ae^x ln x + C
Be^x (1 + x) + C
Ce^x/x^2 + C
De^x/x + C
Answer & Solution
Correct answer: D. e^x/x + C
With f(x) = 1/x, f'(x) = -1/x^2, so the form e^x (f + f') integrates to e^x/x.
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