$\int_0^2\left|x^2+2 x-3\right| d x \text { is equal to }$

Let $f(x) = x^2 + 2x - 3$. We factorize the quadratic expression as $f(x) = (x+3)(x-1)$.

The roots of $f(x) = 0$ are $x = -3$ and $x = 1$. In the interval $[0, 2]$, the function $f(x)$ changes sign at $x = 1$.

For $0 \le x < 1$, $x^2 + 2x - 3 < 0$, so $|x^2 + 2x - 3| = -(x^2 + 2x - 3) = -x^2 - 2x + 3$.

For $1 \le x \le 2$, $x^2 + 2x - 3 \ge 0$, so $|x^2 + 2x - 3| = x^2 + 2x - 3$.

The integral becomes $\int_0^1 (-x^2 - 2x + 3) dx + \int_1^2 (x^2 + 2x - 3) dx$.

Evaluating the first integral: $\left[ -\dfrac{x^3}{3} - x^2 + 3x \right]_0^1 = -\dfrac{1}{3} - 1 + 3 = 2 - \dfrac{1}{3} = \dfrac{5}{3}$.

Evaluating the second integral: $\left[ \dfrac{x^3}{3} + x^2 - 3x \right]_1^2 = \left( \dfrac{8}{3} + 4 - 6 \right) - \left( \dfrac{1}{3} + 1 - 3 \right) = \left( \dfrac{8}{3} - 2 \right) - \left( \dfrac{1}{3} - 2 \right) = \dfrac{8}{3} - \dfrac{1}{3} = \dfrac{7}{3}$.

Summing the results: $\dfrac{5}{3} + \dfrac{7}{3} = \dfrac{12}{3} = 4$.

Answer: 4