(1) Evaluate $ \int_{\minus{}\sqrt{3}}^{\sqrt{3}}( x^2\minus{}1)dx,\  \int_{\minus{}\sqrt{3}}^{\sqrt{3}} (x\minus{}1)^2dx,\  \int_{\minus{}\sqrt{3}}^{\sqrt{3}} (x\plus{}1)^2dx$.

(2) If a linear function $ f(x)$ satifies $ \int_{\minus{}\sqrt{3}}^{\sqrt{3}} (x\minus{}1)f(x)dx\equal{}5\sqrt{3},\ \int_{\minus{}\sqrt{3}}^{\sqrt{3}} (x\plus{}1)f(x)dx\equal{}3\sqrt{3}$, then we have $ f(x)\equal{}\boxed{\ A\ }(x\minus{}1)\plus{}\boxed{\ B\ }(x\plus{}1)$, thus we have $ f(x)\equal{}\boxed{\  C\ }$.
### Part 1: Evaluate the integrals

1. Evaluate \( \int_{-\sqrt{3}}^{\sqrt{3}} (x^2 - 1) \, dx \):

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} (x^2 - 1) \, dx = \int_{-\sqrt{3}}^{\sqrt{3}} x^2 \, dx - \int_{-\sqrt{3}}^{\sqrt{3}} 1 \, dx
   \]

   Since \( x^2 \) is an even function and \( 1 \) is an odd function over the symmetric interval \([-a, a]\):

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} x^2 \, dx = 2 \int_{0}^{\sqrt{3}} x^2 \, dx
   \]

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} 1 \, dx = 2 \int_{0}^{\sqrt{3}} 1 \, dx
   \]

   Now, compute each integral:

   \[
   \int_{0}^{\sqrt{3}} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{\sqrt{3}} = \frac{(\sqrt{3})^3}{3} - \frac{0^3}{3} = \frac{3\sqrt{3}}{3} = \sqrt{3}
   \]

   \[
   \int_{0}^{\sqrt{3}} 1 \, dx = \left[ x \right]_{0}^{\sqrt{3}} = \sqrt{3} - 0 = \sqrt{3}
   \]

   Therefore:

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} x^2 \, dx = 2 \sqrt{3}
   \]

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} 1 \, dx = 2 \sqrt{3}
   \]

   Combining these results:

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} (x^2 - 1) \, dx = 2 \sqrt{3} - 2 \sqrt{3} = 0
   \]

2. Evaluate \( \int_{-\sqrt{3}}^{\sqrt{3}} (x-1)^2 \, dx \):

   \[
   (x-1)^2 = x^2 - 2x + 1
   \]

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} (x-1)^2 \, dx = \int_{-\sqrt{3}}^{\sqrt{3}} x^2 \, dx - 2 \int_{-\sqrt{3}}^{\sqrt{3}} x \, dx + \int_{-\sqrt{3}}^{\sqrt{3}} 1 \, dx
   \]

   Since \( x \) is an odd function over the symmetric interval \([-a, a]\):

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} x \, dx = 0
   \]

   Using the results from the previous integral:

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} (x-1)^2 \, dx = 2 \sqrt{3} - 0 + 2 \sqrt{3} = 4 \sqrt{3}
   \]

3. Evaluate \( \int_{-\sqrt{3}}^{\sqrt{3}} (x+1)^2 \, dx \):

   \[
   (x+1)^2 = x^2 + 2x + 1
   \]

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} (x+1)^2 \, dx = \int_{-\sqrt{3}}^{\sqrt{3}} x^2 \, dx + 2 \int_{-\sqrt{3}}^{\sqrt{3}} x \, dx + \int_{-\sqrt{3}}^{\sqrt{3}} 1 \, dx
   \]

   Using the same results as before:

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} (x+1)^2 \, dx = 2 \sqrt{3} + 0 + 2 \sqrt{3} = 4 \sqrt{3}
   \]

### Part 2: Determine the linear function \( f(x) \)

Given the conditions:

\[
\int_{-\sqrt{3}}^{\sqrt{3}} (x-1) f(x) \, dx = 5\sqrt{3}
\]

\[
\int_{-\sqrt{3}}^{\sqrt{3}} (x+1) f(x) \, dx = 3\sqrt{3}
\]

Assume \( f(x) = A(x-1) + B(x+1) \).

1. Compute \( \int_{-\sqrt{3}}^{\sqrt{3}} (x-1) f(x) \, dx \):

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} (x-1) f(x) \, dx = \int_{-\sqrt{3}}^{\sqrt{3}} (x-1) [A(x-1) + B(x+1)] \, dx
   \]

   \[
   = A \int_{-\sqrt{3}}^{\sqrt{3}} (x-1)^2 \, dx + B \int_{-\sqrt{3}}^{\sqrt{3}} (x-1)(x+1) \, dx
   \]

   \[
   = A \cdot 4\sqrt{3} + B \cdot \int_{-\sqrt{3}}^{\sqrt{3}} (x^2 - 1) \, dx
   \]

   \[
   = A \cdot 4\sqrt{3} + B \cdot 0 = 4A\sqrt{3}
   \]

   Given \( 4A\sqrt{3} = 5\sqrt{3} \):

   \[
   A = \frac{5}{4}
   \]

2. Compute \( \int_{-\sqrt{3}}^{\sqrt{3}} (x+1) f(x) \, dx \):

   \[
   \int_{-\sqrt{3}}^{\sqrt{3}} (x+1) f(x) \, dx = \int_{-\sqrt{3}}^{\sqrt{3}} (x+1) [A(x-1) + B(x+1)] \, dx
   \]

   \[
   = A \int_{-\sqrt{3}}^{\sqrt{3}} (x+1)(x-1) \, dx + B \int_{-\sqrt{3}}^{\sqrt{3}} (x+1)^2 \, dx
   \]

   \[
   = A \cdot \int_{-\sqrt{3}}^{\sqrt{3}} (x^2 - 1) \, dx + B \cdot 4\sqrt{3}
   \]

   \[
   = A \cdot 0 + B \cdot 4\sqrt{3} = 4B\sqrt{3}
   \]

   Given \( 4B\sqrt{3} = 3\sqrt{3} \):

   \[
   B = \frac{3}{4}
   \]

Finally, the linear function \( f(x) \) is:

\[
f(x) = A(x-1) + B(x+1) = \frac{5}{4}(x-1) + \frac{3}{4}(x+1)
\]

Simplify:

\[
f(x) = \frac{5}{4}x - \frac{5}{4} + \frac{3}{4}x + \frac{3}{4} = \frac{8}{4}x - \frac{2}{4} = 2x - \frac{1}{2}
\]

The final answer is \( \boxed{ f(x) = 2x - \frac{1}{2} } \)