Evaluation of Definite Integral with Absolute Value Functions

To evaluate the given integral $I = \int_{-1}^{1} \left( \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} \right) dx,$ we decompose the integrand into its odd and even components: $I = \int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1} dx + \int_{-1}^{1} \frac{|x| + 1}{x^2 + 2|x| + 1} dx$

Since $f(x) = \frac{x^3}{x^2 + 2|x| + 1}$ is an odd function, its integral over the symmetric interval $[-1, 1]$ vanishes. The remaining part $g(x) = \frac{|x| + 1}{x^2 + 2|x| + 1}$ is an even function, allowing us to simplify the expression using the properties of even functions: $I = 2 \int_{0}^{1} \frac{x + 1}{x^2 + 2x + 1} dx$

Factoring the denominator yields $(x + 1)^2$: $I = 2 \int_{0}^{1} \frac{x + 1}{(x + 1)^2} dx = 2 \int_{0}^{1} \frac{1}{x + 1} dx$

Evaluating the definite integral gives: $I = 2 \left[ \log_e (x + 1) \right]_{0}^{1} = 2 (\log_e 2 - \log_e 1) = 2 \log_e 2$

Hence, the correct option is B.