The correct option is A e^x[logx (1/x)]+c >∫>e^x(logx+(1/x^2))dx =>∫>e^x(logx+(1/x) (1/x)+(1/x^2))dx> >=∫>e^x(logx+(1/x) ...

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Integration by Substitution

∫ exlog x+1/x...

Question

$\int e^{x} (l o g x + \frac{1}{x^{2}}) d x =$

e^{x} [l o g x - (\frac{1}{x})] + c

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e^{x} (l n x + \frac{1}{x}) + C

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e^{x} (- l n x - \frac{1}{x}) + C

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e^{x} (- l n x + \frac{1}{x}) + C

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Solution

The correct option is A $e^{x} [l o g x - (\frac{1}{x})] + c$
$\int e^{x} (l o g x + (\frac{1}{x^{2}})) d x = \int e^{x} (l o g x + (\frac{1}{x}) - (\frac{1}{x}) + (\frac{1}{x^{2}})) d x = \int e^{x} (l o g x + (\frac{1}{x})) d x + \int e^{x} ((\frac{- 1}{x}) + (\frac{1}{x^{2}})) d x n o w u s e \int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) = e^{x} l o g x + e^{x} (\frac{- 1}{x}) + c = e^{x} [l o g x - (\frac{1}{x})] + c$

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