Evaluate  $ \int_0^1 (1 \plus{} x \plus{} x^2 \plus{} \cdots \plus{} x^{n \minus{} 1})\{1 \plus{} 3x \plus{} 5x^2 \plus{} \cdots \plus{} (2n \minus{} 3)x^{n \minus{} 2} \plus{} (2n \minus{} 1)x^{n \minus{} 1}\}\ dx.$
1. Define the functions \( a_n \) and \( b_n \) as follows:
   \[
   a_n = 1 + x + x^2 + \cdots + x^{n-1}
   \]
   \[
   b_n = 1 + 3x + 5x^2 + \cdots + (2n-3)x^{n-2} + (2n-1)x^{n-1}
   \]

2. We need to evaluate the integral:
   \[
   f(n) = \int_0^1 a_n b_n \, dx
   \]

3. Observe that:
   \[
   a_{n+1} = a_n + x^n
   \]
   \[
   b_{n+1} = b_n + (2n+1)x^n
   \]

4. Therefore:
   \[
   a_{n+1} b_{n+1} = (a_n + x^n)(b_n + (2n+1)x^n)
   \]
   \[
   = a_n b_n + (2n+1)x^{2n} + (2n+1)a_n x^n + b_n x^n
   \]

5. Simplify the expression:
   \[
   a_{n+1} b_{n+1} = a_n b_n + (2n+1)x^{2n} + [(2n+1)a_n + b_n]x^n
   \]

6. Notice that:
   \[
   [(2n+1)a_n + b_n]x^n = 2[(n+1) + (n+2)x + \cdots + 2nx^{n-1}]x^n
   \]

7. Thus:
   \[
   a_{n+1} b_{n+1} - a_n b_n = (x^{2n+1})' + 2(x^{n+1} + x^{n+2} + \cdots + x^{2n})'
   \]

8. Integrate both sides from 0 to 1:
   \[
   \int_0^1 (a_{n+1} b_{n+1} - a_n b_n) \, dx = \int_0^1 (x^{2n+1})' \, dx + 2 \int_0^1 (x^{n+1} + x^{n+2} + \cdots + x^{2n})' \, dx
   \]

9. Evaluate the integrals:
   \[
   \int_0^1 (x^{2n+1})' \, dx = x^{2n+1} \Big|_0^1 = 1
   \]
   \[
   2 \int_0^1 (x^{n+1} + x^{n+2} + \cdots + x^{2n})' \, dx = 2 \left( \sum_{k=n+1}^{2n} \int_0^1 x^k \, dx \right)
   \]
   \[
   = 2 \left( \sum_{k=n+1}^{2n} \frac{1}{k+1} \right)
   \]

10. Combine the results:
    \[
    f(n+1) - f(n) = 2n + 1
    \]

11. Given that \( f(1) = 1 \), we can solve the recurrence relation:
    \[
    f(n) = n^2
    \]

The final answer is \( \boxed{n^2} \).