int_{-1}^{1} frac{x^{3} + |x| + 3}{x^{2} + 4| | vedclass

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(B) Let $I = \int_{-1}^{1} \frac{x^{3} + |x| + 3}{x^{2} + 4|x| + 3} dx$. Since $f(x) = \frac{x^{3} + |x| + 3}{x^{2} + 4|x| + 3}$,we can split the inte

Vedclass · Accepted Answer

$-\frac{4}{\pi} \int_{0}^{\frac{\pi}{2}} \log(\sin 	heta) d	heta$

Vedclass · Answer

(B) Let $I = \int_{-1}^{1} \frac{x^{3} + |x| + 3}{x^{2} + 4|x| + 3} dx$. Since $f(x) = \frac{x^{3} + |x| + 3}{x^{2} + 4|x| + 3}$,we can split the integral: $I = \int_{-1}^{1} \frac{x^{3}}{x^{2} + 4|x| + 3} dx + \int_{-1}^{1} \frac{|x| + 3}{x^{2} + 4|x| + 3} dx$. The first part is an odd function,so its integral over $[-1, 1]$ is $0$. For the second part,since $|x| + 3 = (|x| + 1)(|x| + 3)$ is not correct,we note $x^2 = |x|^2$. So,$\frac{|x| + 3}{|x|^2 + 4|x| + 3} = \frac{|x| + 3}{(|x| + 1)(|x| + 3)} = \frac{1}{|x| + 1}$. Thus,$I = 2 \int_{0}^{1} \frac{1}{x + 1} dx = 2 [\ln(x + 1)]_{0}^{1} = 2 \ln 2$. Using the standard integral $\int_{0}^{\frac{\pi}{2}} \log(\sin \alpha) d\alpha = -\frac{\pi}{2} \ln 2$,we see that $- \frac{4}{\pi} (-\frac{\pi}{2} \ln 2) = 2 \ln 2$. Thus,the correct option is $B$.

$\int_{-1}^{1} \frac{x^{3} + |x| + 3}{x^{2} + 4|x| + 3} dx$ is equal to -

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