The value of the integral int_{-1}^1 left( fr | vedclass

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(B) Let $I = \int_{-1}^1 \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} dx$. Since the denominator $x^2 + 2|x| + 1 = (|x| + 1)^2$ is an even function,we can spl

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$2 \log_e 2$

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(B) Let $I = \int_{-1}^1 \frac{x^3 + |x| + 1}{x^2 + 2|x| + 1} dx$. Since the denominator $x^2 + 2|x| + 1 = (|x| + 1)^2$ is an even function,we can split the integral into two parts: $I = \int_{-1}^0 \frac{x^3 - x + 1}{(|x| + 1)^2} dx + \int_0^1 \frac{x^3 + x + 1}{(|x| + 1)^2} dx$. For the first part,let $x = -t$,then $dx = -dt$. As $x$ goes from $-1$ to $0$,$t$ goes from $1$ to $0$. $\int_1^0 \frac{-t^3 + t + 1}{(t + 1)^2} (-dt) = \int_0^1 \frac{-t^3 + t + 1}{(t + 1)^2} dt$. Thus,$I = \int_0^1 \frac{x^3 + x + 1 - x^3 + x + 1}{(x + 1)^2} dx = \int_0^1 \frac{2x + 2}{(x + 1)^2} dx = \int_0^1 \frac{2(x + 1)}{(x + 1)^2} dx = \int_0^1 \frac{2}{x + 1} dx$. $I = 2 [\log_e(x + 1)]_0^1 = 2(\log_e 2 - \log_e 1) = 2 \log_e 2$.

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

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