Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \mathrm{e}^{x}\left(\log x+\frac{1}{x^{2}}\right) d x$

Solution:

Let $I=\int e^{x}\left(\log x+\frac{1}{x^{2}}\right) d x$

$=\int e^{x}\left(\log x+\frac{1}{x}-\frac{1}{x}+\frac{1}{x^{2}}\right) d x$

$=\int e^{x}\left(\log x-\frac{1}{x}\right) d x+\int e^{x}\left(\frac{1}{x}+\frac{1}{x^{2}}\right) d x$

Using integration by parts,

$=e^{x}\left(\log x-\frac{1}{x}\right)-\int e^{x} \frac{d}{d x}\left(\log x-\frac{1}{x}\right) d x+\int e^{x}\left(\frac{1}{x}+\frac{1}{x^{2}}\right) d x$

$=e^{x}\left(\log x-\frac{1}{x}\right)-\int e^{x}\left(\frac{1}{x}+\frac{1}{x^{2}}\right) d x+\int e^{x}\left(\frac{1}{x}+\frac{1}{x^{2}}\right) d x$

$=e^{x}\left(\log x-\frac{1}{x}\right)+c$

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