Step 1: Write out the integral.
We are given $\int_1^3 x(x^2-1)^{1/n}\,dx = 6$ and need to find the positive integer $n$.
Step 2: Use the substitution $t = x^2 - 1$.
Let $t = x^2 - 1$, so $dt = 2x\,dx$, i.e., $x\,dx = dt/2$. When $x = 1$, $t = 0$; when $x = 3$, $t = 8$.
Step 3: Transform the integral.
$\int_1^3 x(x^2-1)^{1/n}\,dx = \int_0^8 t^{1/n} \cdot \frac{dt}{2} = \frac{1}{2}\int_0^8 t^{1/n}\,dt$.
Step 4: Evaluate the integral.
$\frac{1}{2}\int_0^8 t^{1/n}\,dt = \frac{1}{2}\cdot\left[\frac{t^{1/n+1}}{1/n+1}\right]_0^8 = \frac{1}{2}\cdot\frac{n}{n+1}\cdot 8^{(n+1)/n} = \frac{n}{2(n+1)}\cdot 8^{(n+1)/n}$.
Step 5: Set equal to 6 and test $n = 3$.
$\frac{n}{2(n+1)}\cdot 8^{(n+1)/n} = 6$. Try $n=3$: $8^{4/3} = (8^{1/3})^4 = 2^4 = 16$. Then $\frac{3}{2\cdot 4}\cdot 16 = \frac{3\cdot 16}{8} = \frac{48}{8} = 6$. This works!
Step 6: State the final answer.
\[\boxed{n = 3}\]