Find $\int x \cos x \, dx$

$x \log x - x + C$

The correct answer is Option (2) → $x \log x - x + C$

To start with, we are unable to guess a function whose derivative is $\log x$. We take $\log x$ as the first function and the constant function $1$ as the second function. Then, the integral of the second function is $x$.

Hence,

$\int (\log x \cdot 1) \, dx = \log x \int 1 \, dx - \int \left[ \frac{d}{dx} (\log x) \int 1 \, dx \right] dx$

$= (\log x) \cdot x - \int \frac{1}{x} \cdot x \, dx = x \log x - x + C.$