intlimits_{ - 1}^1 {frac{{{x^3} + |x| + 1}}{{ | vedclass

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(B) Let $I = \int\limits_{ - 1}^1 {\frac{{{x^3} + |x| + 1}}{{{x^2} + 2|x| + 1}}} dx$. Since the denominator ${x^2} + 2|x| + 1 = (|x| + 1)^2$ is an eve

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$a = 2; b = 0$

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(B) Let $I = \int\limits_{ - 1}^1 {\frac{{{x^3} + |x| + 1}}{{{x^2} + 2|x| + 1}}} dx$. Since the denominator ${x^2} + 2|x| + 1 = (|x| + 1)^2$ is an even function,we can split the integral into odd and even parts. $I = \int\limits_{ - 1}^1 {\frac{{{x^3}}}{{(|x| + 1)^2}}} dx + \int\limits_{ - 1}^1 {\frac{{|x| + 1}}{{(|x| + 1)^2}}} dx$. The first part $\int\limits_{ - 1}^1 {\frac{{{x^3}}}{{(|x| + 1)^2}}} dx$ is an integral of an odd function over a symmetric interval $[-1, 1]$,so it equals $0$. Thus,$I = \int\limits_{ - 1}^1 {\frac{1}{{|x| + 1}}} dx$. Since $\frac{1}{{|x| + 1}}$ is an even function,$I = 2 \int\limits_0^1 {\frac{1}{{x + 1}}} dx$. $I = 2 [\ln(x + 1)]_0^1 = 2(\ln 2 - \ln 1) = 2 \ln 2$. Comparing $2 \ln 2$ with $a \ln 2 + b$,we get $a = 2$ and $b = 0$.

$\int\limits_{ - 1}^1 {\frac{{{x^3} + |x| + 1}}{{{x^2} + 2|x| + 1}}} dx = a \ln 2 + b$,then:

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