$\int_{-1}^1 \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1}dx$ is equal to

The integral to evaluate is $\int_{-1}^1 \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1}dx$

The integrand contains absolute values and the limits cross $x = 0$.

---

The denominator can be simplified:

$x^2 + 2|x| + 1 = (|x|)^2 + 2|x| + 1$

$= (|x| + 1)^2$

---

Let $f(x) = \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1}$

For $x > 0$:

$f(x) = \frac{x^3 + x + 1}{(x+1)^2}$

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

Adding these:

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

$= \frac{2x + 2}{(x+1)^2}$

$= \frac{2(x+1)}{(x+1)^2}$

$= \frac{2}{x+1}$

---

Using the symmetry property for integrals over $[-a, a]$:

$\int_{-1}^1 f(x)dx = \int_0^1 [f(x) + f(-x)]dx$

Therefore:

$\int_{-1}^1 \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1}dx = \int_0^1 \frac{2}{x+1}dx$

---

Evaluating the integral:

$\int_0^1 \frac{2}{x+1}dx = 2\int_0^1 \frac{1}{x+1}dx$

$= 2[\ln(x+1)]_0^1$

$= 2[\ln(2) - \ln(1)]$

$= 2\ln(2)$

$= 2\log_e 2$

Therefore, the answer is $2\log_e 2$.