The value of the integral $$\displaystyle\int_{-1}^{1} \left(\dfrac{x^3 + |x| + 1}{x^2 + 2|x| + 1}\right) dx$$ is equal to :

$$I = \int_{-1}^{1} \left( \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} \right) dx$$

$$I = \int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} \, dx + \int_{-1}^{1} \frac{|x| + 1}{x^2 + 2|x| + 1} \, dx$$

$$f_1(-x) = \frac{(-x)^3}{(-x)^2 + 2|-x| + 1} = \frac{-x^3}{x^2 + 2|x| + 1} = -f_1(x)$$

Since $$f_1(x)$$ is an odd function, its integral from $$-1$$ to $$1$$ is zero: $$\int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} \, dx = 0$$

$$I = \int_{-1}^{1} \frac{|x| + 1}{x^2 + 2|x| + 1} \, dx = 2 \int_{0}^{1} \frac{|x| + 1}{x^2 + 2|x| + 1} \, dx$$

$$I = 2 \int_{0}^{1} \frac{x + 1}{x^2 + 2x + 1} \, dx$$

$$I = 2 \int_{0}^{1} \frac{x + 1}{(x + 1)^2} \, dx$$

$$I = 2 \int_{0}^{1} \frac{1}{x + 1} \, dx$$

$$I = 2 \Big[ \log_e(x + 1) \Big]_0^1$$

$$I = 2 \left[ \log_e 2 - \log_e 1 \right]$$

$$I = 2 \log_e 2$$