Question

Evaluate the following integrals:

${\int }_{-2}^2\frac{3x^3+2\left|x\right|+1}{x^2+\left|x\right|+1}dx$

Answer 1

Let I = ${\int }_{-2}^2\frac{3x^3+2\left|x\right|+1}{x^2+\left|x\right|+1}dx$

$$\begin{gathered} ={\int }_{-2}^2\frac{3x^3}{x^2+\left|x\right|+1}dx+{\int }_{-2}^2\frac{2\left|x\right|+1}{x^2+\left|x\right|+1}dx \\ =I_1+I_2 \end{gathered}$$

Consider $f\left(x\right)={\int }_{-2}^2\frac{3x^3}{x^2+\left|x\right|+1}dx$.

$f\left(-x\right)={\int }_{-2}^2\frac{3{\left(-x\right)}^3}{{\left(-x\right)}^2+\left|-x\right|+1}dx={\int }_{-2}^2\frac{-3x^3}{x^2+\left|x\right|+1}dx=-{\int }_{-2}^2\frac{3x^3}{x^2+\left|x\right|+1}dx=-f\left(x\right)$

$\therefore I_1={\int }_{-2}^2\frac{3x^3}{x^2+\left|x\right|+1}dx=0\left[{\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{ll}2{\int }_0^{a}f\left(x\right)dx,& \mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,& \mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right.\right]$

Now, consider $g\left(x\right)={\int }_{-2}^2\frac{2\left|x\right|+1}{x^2+\left|x\right|+1}dx$.

$g\left(-x\right)={\int }_{-2}^2\frac{2\left|-x\right|+1}{{\left(-x\right)}^2+\left|-x\right|+1}dx={\int }_{-2}^2\frac{2\left|x\right|+1}{x^2+\left|x\right|+1}dx=g\left(x\right)$

$$\begin{gathered} \therefore I_2={\int }_{-2}^2\frac{2\left|x\right|+1}{x^2+\left|x\right|+1}dx \\ =2{\int }_0^2\frac{2\left|x\right|+1}{x^2+\left|x\right|+1}dx\left[{\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{ll}2{\int }_0^{a}f\left(x\right)dx,& \mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,& \mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right.\right] \\ =2{\int }_0^2\frac{2x+1}{x^2+x+1}dx\left[\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}\right.\right] \\ =2\times {\overline{)\log \left(x^2+x+1\right)}}_0^2\left[\int \frac{f\text{'}\left(x\right)}{f\left(x\right)}dx=\log f\left(x\right)+C\right] \\ =2\times \left(\log 7-\log 1\right) \\ =2\times \left(\log 7-0\right) \\ =2\log 7 \end{gathered}$$

$\therefore I=I_1+I_2=0+2\log 7=2\log 7$