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Question

Evaluate the following integrals: ${\int }_{-2}^{2}\frac{3{x}^{3}+2\left|x\right|+1}{{x}^{2}+\left|x\right|+1}dx$

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Solution

Let I = ${\int }_{-2}^{2}\frac{3{x}^{3}+2\left|x\right|+1}{{x}^{2}+\left|x\right|+1}dx$ $={\int }_{-2}^{2}\frac{3{x}^{3}}{{x}^{2}+\left|x\right|+1}dx+{\int }_{-2}^{2}\frac{2\left|x\right|+1}{{x}^{2}+\left|x\right|+1}dx\phantom{\rule{0ex}{0ex}}={I}_{1}+{I}_{2}$ Consider $f\left(x\right)={\int }_{-2}^{2}\frac{3{x}^{3}}{{x}^{2}+\left|x\right|+1}dx$. $f\left(-x\right)={\int }_{-2}^{2}\frac{3{\left(-x\right)}^{3}}{{\left(-x\right)}^{2}+\left|-x\right|+1}dx={\int }_{-2}^{2}\frac{-3{x}^{3}}{{x}^{2}+\left|x\right|+1}dx=-{\int }_{-2}^{2}\frac{3{x}^{3}}{{x}^{2}+\left|x\right|+1}dx=-f\left(x\right)$ $\therefore {I}_{1}={\int }_{-2}^{2}\frac{3{x}^{3}}{{x}^{2}+\left|x\right|+1}dx=0\left[{\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{ll}2{\int }_{0}^{a}f\left(x\right)dx,& \mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,& \mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right\\right]$ Now, consider $g\left(x\right)={\int }_{-2}^{2}\frac{2\left|x\right|+1}{{x}^{2}+\left|x\right|+1}dx$. $g\left(-x\right)={\int }_{-2}^{2}\frac{2\left|-x\right|+1}{{\left(-x\right)}^{2}+\left|-x\right|+1}dx={\int }_{-2}^{2}\frac{2\left|x\right|+1}{{x}^{2}+\left|x\right|+1}dx=g\left(x\right)$ $\therefore {I}_{2}={\int }_{-2}^{2}\frac{2\left|x\right|+1}{{x}^{2}+\left|x\right|+1}dx\phantom{\rule{0ex}{0ex}}=2{\int }_{0}^{2}\frac{2\left|x\right|+1}{{x}^{2}+\left|x\right|+1}dx\left[{\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{ll}2{\int }_{0}^{a}f\left(x\right)dx,& \mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,& \mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right\\right]\phantom{\rule{0ex}{0ex}}=2{\int }_{0}^{2}\frac{2x+1}{{x}^{2}+x+1}dx\left[\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}\right\\right]\phantom{\rule{0ex}{0ex}}=2×{\overline{)\mathrm{log}\left({x}^{2}+x+1\right)}}_{0}^{2}\left[\int \frac{f\text{'}\left(x\right)}{f\left(x\right)}dx=\mathrm{log}f\left(x\right)+C\right]\phantom{\rule{0ex}{0ex}}=2×\left(\mathrm{log}7-\mathrm{log}1\right)\phantom{\rule{0ex}{0ex}}=2×\left(\mathrm{log}7-0\right)\phantom{\rule{0ex}{0ex}}=2\mathrm{log}7$ $\therefore I={I}_{1}+{I}_{2}=0+2\mathrm{log}7=2\mathrm{log}7$

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