I=∫exlnx+1x2dx==∫exlnx-1x+1x+1x2dx==∫exfx+f'xdx where fx=lnx-1x ⇒f'x=1x+1x2=exfx+c=exlnx-1x+c

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$\int e^{x} [\ln x + \frac{1}{x^{2}}] d x =$

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$e^{x} \ln x + C$

$e^{x} (\ln x - \frac{1}{x}) + c$

$e^{x} [\ln x - \frac{1}{x^{2}}] + c$

$\frac{e^{x}}{x^{2}} + c$

answer is B.

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Detailed Solution

$I = \int e^{x} [\ln x + \frac{1}{x^{2}}] d x =$

$=$ $\int e^{x} [(\ln x - \frac{1}{x}) + (\frac{1}{x} + \frac{1}{x^{2}})] d x =$

$= \int e^{x} [f (x) + f^{'} (x)] d x$
$w h e r e f (x) = \ln x - \frac{1}{x} \Rightarrow f^{'} (x) = \frac{1}{x} + \frac{1}{x^{2}}$

$= e^{x} f (x) + c$

$= e^{x} (\ln x - \frac{1}{x}) + c$

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