Question:
Evaluate the following integrals:
$\int e^{x}\left(\log x+\frac{1}{x}\right) d x$
Solution:
Let $I=\int e^{x}\left(\log x+\frac{1}{x}\right) d x$
We know that
$\int e^{x}\left\{f(x)+f^{\prime}(x)\right\}=e^{x} f(x)+c$
Here,
$f(x)=\log x ; f^{\prime}(x)=\frac{1}{x}$
$\int e^{x}\left(\log x+\frac{1}{x}\right) d x=e^{x} \log x+c$
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