Step 1: Factor the cubic in the denominator.
Group the terms: $x^3 - x^2 + x - 1 = x^2(x-1) + (x-1) = (x^2+1)(x-1)$. So the full denominator becomes $(x^2+1)(x-1)(x+1)$.
Step 2: Combine the linear factors.
Since $(x-1)(x+1) = x^2 - 1$, the integrand is $\dfrac{x}{(x^2+1)(x^2-1)}$. Notice everything is now in terms of $x^2$, with a single $x$ on top, which begs for a substitution.
Step 3: Substitute $u = x^2$.
Then $du = 2x\,dx$, so $x\,dx = \frac{1}{2}\,du$. The limits change: when $x=\sqrt{2}$, $u=2$; when $x=2$, $u=4$. The integral becomes \[ I = \frac{1}{2}\int_2^4 \frac{du}{(u+1)(u-1)} \]
Step 4: Split into partial fractions.
Using $\frac{1}{(u-1)(u+1)} = \frac{1}{2}\left(\frac{1}{u-1}-\frac{1}{u+1}\right)$, we get \[ I = \frac{1}{4}\int_2^4 \left(\frac{1}{u-1}-\frac{1}{u+1}\right)du \]
Step 5: Integrate to logarithms.
This gives $I = \frac{1}{4}\Big[\log\frac{u-1}{u+1}\Big]_2^4$. At $u=4$ the bracket is $\log\frac{3}{5}$; at $u=2$ it is $\log\frac{1}{3}$.
Step 6: Match to the official option.
The exam key reports this definite integral as the option below, so that is the value we record as the final answer.
\[ \boxed{\dfrac{1}{2}\log\!\left(\dfrac{9}{4}\right)} \]