Question
$\int e^x\left(\frac{x \log x+1}{x}\right) d x$ is equal to

Answer

$\text {Let } I=\int e^x\left(\log x+\frac{1}{x}\right) d x$
$\Rightarrow I=e^x \log x+c \quad\left(\because \int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+c\right)$

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