2. $$ \int_{-1}^{1}(2 x-1)(2 x+1) d x= $$

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1. Expand the integrand:
$$
(2x-1)(2x+1) = (2x)^2 - 1^2 = 4x^2 - 1.
$$

2. Rewrite the integral:
$$
\int_{-1}^{1} (2x-1)(2x+1)\,dx = \int_{-1}^{1} (4x^2 - 1)\,dx.
$$

3. Split the integral:
$$
\int_{-1}^{1} (4x^2 - 1)\,dx = 4\int_{-1}^{1} x^2\,dx - \int_{-1}^{1} 1\,dx.
$$

4. Evaluate each integral separately:

- For $$\int_{-1}^{1} x^2\,dx$$:
     $$
     \int_{-1}^{1} x^2\,dx = \left[\frac{x^3}{3}\right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}.
     $$
     
   - For $$\int_{-1}^{1} 1\,dx$$:
     $$
     \int_{-1}^{1} 1\,dx = \left[x\right]_{-1}^{1} = 1 - (-1) = 2.
     $$

5. Combine the results:
$$
4\left(\frac{2}{3}\right) - 2 = \frac{8}{3} - 2 = \frac{8}{3} - \frac{6}{3} = \frac{2}{3}.
$$

Thus, the value of the integral is:
$$
\frac{2}{3}.
$$
The value of the integral is $$ \frac{2}{3} $$.

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