The value of the integral int_{-1}^1 frac{|x+ | vedclass

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(B) Let $I = \int_{-1}^1 \frac{|x+2|}{x+2} \, dx$. \ Since the interval of integration is $[-1, 1]$, we have $x \geq -1$. \ This implies $x+2 \geq 1

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$2$

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(B) Let $I = \int_{-1}^1 \frac{|x+2|}{x+2} \, dx$. \ Since the interval of integration is $[-1, 1]$, we have $x \geq -1$. \ This implies $x+2 \geq 1$, so $x+2$ is always positive in the given interval. \ Therefore, $|x+2| = x+2$. \ Substituting this into the integral, we get: \ $I = \int_{-1}^1 \frac{x+2}{x+2} \, dx = \int_{-1}^1 1 \, dx$. \ Evaluating the integral: \ $I = [x]_{-1}^1 = 1 - (-1) = 2$.

The value of the integral $\int_{-1}^1 \frac{|x+2|}{x+2} \, dx$ is

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