int_{-1}^1 frac{x^3+|x|+1}{x^2+2|x|+1} dx is | vedclass

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(B) Let $I = \int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} dx$. We can split the integral into two parts: $I = \int_{-1}^1 \frac{x^3}{x^2+2|x|+1} dx + \int

Vedclass · Accepted Answer

$2 \log 2$

Vedclass · Answer

(B) Let $I = \int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} dx$. We can split the integral into two parts: $I = \int_{-1}^1 \frac{x^3}{x^2+2|x|+1} dx + \int_{-1}^1 \frac{|x|+1}{x^2+2|x|+1} dx$. Let $f(x) = \frac{x^3}{x^2+2|x|+1}$. Since $f(-x) = \frac{(-x)^3}{(-x)^2+2|-x|+1} = -\frac{x^3}{x^2+2|x|+1} = -f(x)$, the function is odd. Therefore, $\int_{-1}^1 f(x) dx = 0$. Now, consider the second part $I_2 = \int_{-1}^1 \frac{|x|+1}{x^2+2|x|+1} dx$. Since the integrand is an even function, $I_2 = 2 \int_0^1 \frac{x+1}{x^2+2x+1} dx$. Simplifying the integrand: $I_2 = 2 \int_0^1 \frac{x+1}{(x+1)^2} dx = 2 \int_0^1 \frac{1}{x+1} dx$. Evaluating the integral: $I_2 = 2 [\ln |x+1|]_0^1 = 2 (\ln 2 - \ln 1) = 2 \ln 2$. Thus, $I = 0 + 2 \ln 2 = 2 \ln 2$.

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

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