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

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

We can split the integral into two parts:

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

The first integrand $f(x) = \dfrac{x^3}{x^2 + 2|x| + 1}$ is an odd function since $f(-x) = -f(x)$. Therefore, its integral over the symmetric interval $[-1, 1]$ is zero.

The second integrand $g(x) = \dfrac{|x| + 1}{x^2 + 2|x| + 1}$ is an even function since $g(-x) = g(x)$. Therefore, its integral over $[-1, 1]$ is twice the integral over $[0, 1]$.

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

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

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

$I = 2 [\log_e(x + 1)]_{0}^{1}$

$I = 2(\log_e 2 - \log_e 1) = 2\log_e 2$

Answer: $2\log_e 2$