The integral $\int e^x\left(\log x+\frac{1}{x^2}\right) d x$ equals for some arbitrary constant k

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

The correct answer is Option (2) → $e^x\left(\log x-\frac{1}{x}\right)+k$

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

$=\int e^x\left(\underset{f(x)}{\underbrace{\log x}}+\underset{f'(x)}{\underbrace{\frac{1}{x}}}\right) d x+\int e^x\left(\underset{g(x)}{\underbrace{-\frac{1}{x}}}+\underset{g'(x)}{\underbrace{\frac{1}{x^2}}}\right)dx$

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