Let $I_n=\frac{1}{n+1}\int_0^{\pi} x(\sin nx+n\pi\cos nx)dx\ \ (n=1,\ 2,\ \cdots).$

Answer the questions below.

(1) Find $I_n.$

(2) Find $\sum_{n=1}^{\infty} I_n.$
1. **Finding \( I_n \):**

   Given:
   \[
   I_n = \frac{1}{n+1} \int_0^{\pi} x (\sin(nx) + n\pi \cos(nx)) \, dx
   \]

   We can split the integral into two parts:
   \[
   I_n = \frac{1}{n+1} \left( \int_0^{\pi} x \sin(nx) \, dx + n\pi \int_0^{\pi} x \cos(nx) \, dx \right)
   \]

   Let's evaluate each integral separately.

   **First Integral:**
   \[
   \int_0^{\pi} x \sin(nx) \, dx
   \]

   Using integration by parts, let \( u = x \) and \( dv = \sin(nx) \, dx \). Then \( du = dx \) and \( v = -\frac{1}{n} \cos(nx) \).

   \[
   \int_0^{\pi} x \sin(nx) \, dx = \left. -\frac{x}{n} \cos(nx) \right|_0^{\pi} + \frac{1}{n} \int_0^{\pi} \cos(nx) \, dx
   \]

   Evaluating the boundary term:
   \[
   \left. -\frac{x}{n} \cos(nx) \right|_0^{\pi} = -\frac{\pi}{n} \cos(n\pi) + 0 = -\frac{\pi}{n} (-1)^n
   \]

   Now, evaluate the remaining integral:
   \[
   \frac{1}{n} \int_0^{\pi} \cos(nx) \, dx = \frac{1}{n} \left. \frac{\sin(nx)}{n} \right|_0^{\pi} = \frac{1}{n^2} (\sin(n\pi) - \sin(0)) = 0
   \]

   Therefore:
   \[
   \int_0^{\pi} x \sin(nx) \, dx = -\frac{\pi}{n} (-1)^n = \frac{\pi}{n} (-1)^n
   \]

   **Second Integral:**
   \[
   n\pi \int_0^{\pi} x \cos(nx) \, dx
   \]

   Using integration by parts again, let \( u = x \) and \( dv = \cos(nx) \, dx \). Then \( du = dx \) and \( v = \frac{1}{n} \sin(nx) \).

   \[
   \int_0^{\pi} x \cos(nx) \, dx = \left. \frac{x}{n} \sin(nx) \right|_0^{\pi} - \frac{1}{n} \int_0^{\pi} \sin(nx) \, dx
   \]

   Evaluating the boundary term:
   \[
   \left. \frac{x}{n} \sin(nx) \right|_0^{\pi} = \frac{\pi}{n} \sin(n\pi) - 0 = 0
   \]

   Now, evaluate the remaining integral:
   \[
   \frac{1}{n} \int_0^{\pi} \sin(nx) \, dx = \frac{1}{n} \left. -\frac{\cos(nx)}{n} \right|_0^{\pi} = -\frac{1}{n^2} (\cos(n\pi) - \cos(0)) = -\frac{1}{n^2} ((-1)^n - 1)
   \]

   Therefore:
   \[
   n\pi \int_0^{\pi} x \cos(nx) \, dx = n\pi \left( 0 - \frac{1}{n^2} ((-1)^n - 1) \right) = -\frac{\pi}{n} ((-1)^n - 1)
   \]

   Combining both integrals:
   \[
   I_n = \frac{1}{n+1} \left( \frac{\pi}{n} (-1)^n - \frac{\pi}{n} ((-1)^n - 1) \right) = \frac{1}{n+1} \left( \frac{\pi}{n} \right) = \frac{\pi}{n(n+1)}
   \]

2. **Finding \( \sum_{n=1}^{\infty} I_n \):**

   \[
   \sum_{n=1}^{\infty} I_n = \sum_{n=1}^{\infty} \frac{\pi}{n(n+1)}
   \]

   We can use partial fraction decomposition:
   \[
   \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}
   \]

   Therefore:
   \[
   \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right)
   \]

   This is a telescoping series:
   \[
   \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) = 1
   \]

   Therefore:
   \[
   \sum_{n=1}^{\infty} I_n = \pi \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \pi \cdot 1 = \pi
   \]

The final answer is \( \boxed{ \pi } \)