(1) Evaluate $ \int_0^1 xe^{x^2}dx$.

(2) Let $ I_n\equal{}\int_0^1 x^{2n\minus{}1}e^{x^2}dx$. Express $ I_{n\plus{}1}$ in terms of $ I_n$.
### Part 1: Evaluate \( \int_0^1 xe^{x^2} \, dx \)

1. **Substitution**: Let \( x^2 = t \). Then \( 2x \, dx = dt \) or \( x \, dx = \frac{1}{2} \, dt \).

2. **Change of Limits**: When \( x = 0 \), \( t = 0 \). When \( x = 1 \), \( t = 1 \).

3. **Rewrite the Integral**:
   \[
   \int_0^1 xe^{x^2} \, dx = \int_0^1 e^t \cdot \frac{1}{2} \, dt = \frac{1}{2} \int_0^1 e^t \, dt
   \]

4. **Evaluate the Integral**:
   \[
   \frac{1}{2} \int_0^1 e^t \, dt = \frac{1}{2} \left[ e^t \right]_0^1 = \frac{1}{2} (e^1 - e^0) = \frac{1}{2} (e - 1)
   \]

### Part 2: Express \( I_{n+1} \) in terms of \( I_n \)

1. **Definition**: Let \( I_n = \int_0^1 x^{2n-1} e^{x^2} \, dx \).

2. **Expression for \( I_{n+1} \)**:
   \[
   I_{n+1} = \int_0^1 x^{2n+1} e^{x^2} \, dx = \int_0^1 x^{2n} \cdot x e^{x^2} \, dx
   \]

3. **Integration by Parts**: Let \( u = x^{2n} \) and \( dv = x e^{x^2} \, dx \). Then \( du = 2n x^{2n-1} \, dx \) and \( v = \frac{e^{x^2}}{2} \).

4. **Apply Integration by Parts**:
   \[
   \int u \, dv = uv - \int v \, du
   \]
   \[
   I_{n+1} = \left. x^{2n} \cdot \frac{e^{x^2}}{2} \right|_0^1 - \int_0^1 \frac{e^{x^2}}{2} \cdot 2n x^{2n-1} \, dx
   \]

5. **Evaluate the Boundary Term**:
   \[
   \left. x^{2n} \cdot \frac{e^{x^2}}{2} \right|_0^1 = \left( 1^{2n} \cdot \frac{e^1}{2} \right) - \left( 0^{2n} \cdot \frac{e^0}{2} \right) = \frac{e}{2} - 0 = \frac{e}{2}
   \]

6. **Simplify the Integral**:
   \[
   I_{n+1} = \frac{e}{2} - n \int_0^1 x^{2n-1} e^{x^2} \, dx = \frac{e}{2} - n I_n
   \]

The final answer is \( \boxed{\frac{1}{2}(e - 1)} \) for part 1 and \( \boxed{I_{n+1} = \frac{e}{2} - n I_n} \) for part 2.