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