Step 1: Choose a method:
The integrand does not fit a simple power rule. A good guess for the antiderivative is \( \dfrac{x}{\log x} \), and we prove it by differentiating.
Step 2: Differentiate the guess:
Let \( g(x) = \dfrac{x}{\log x} \). By the quotient rule, \( g^{\prime}(x) = \dfrac{\log x \cdot 1 - x \cdot \dfrac1x}{(\log x)^{2}} \).
Step 3: Simplify:
\[ g^{\prime}(x) = \dfrac{\log x - 1}{(\log x)^{2}} = \dfrac{1}{\log x} - \dfrac{1}{(\log x)^{2}} \]
Step 4: Conclude:
This is exactly the integrand. So the integral is \( \dfrac{x}{\log x} + C \).
Step 5: Check the other options:
Differentiating \( \dfrac{1}{\log x} \) gives \( -\dfrac{1}{x(\log x)^{2}} \), not the integrand. Differentiating \( x^{2}\log x \) gives \( 2x\log x + x \). Differentiating option (D) gives \( \dfrac{2}{x(\log x)^{3}} \). None matches.
Final Answer:
The integral is \( \dfrac{x}{\log x} + C \), option (A).
\[ \boxed{\dfrac{x}{\log x} + C} \]