The correct option is A e^xlog |x|+c Now, ∫e^xx(1+xlog |x|) dx =∫e^xx dx+ ∫e^xlog |x| dx =e^xlog |x| ∫ e^xlog |x|dx+ ∫e^xlog |x| d ...

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Integration by Parts

∫exx1+xlog |x...

Question

$\int \frac{e^{x}}{x} (1 + x log | x |) d x =$

e^{x} log | x | + c

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e^{x} log | x | + e^{x} + c

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e^{x} log | s | - x + c

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none of these

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Solution

The correct option is A $e^{x} log | x | + c$
Now,
$\int \frac{e^{x}}{x} (1 + x log | x |) d x$
$= \int \frac{e^{x}}{x} d x +$ $\int e^{x} log | x | d x$
$= e^{x} log | x | - \int e^{x} log | x | d x +$ $\int e^{x} log | x | d x + c$
$= e^{x} log | x | + c$ . [ Where $c$ is integrating constant]

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