Question

Evaluate ${\int }_{-1}^1\frac{x+|x|+1}{x^2+2|x|+1}dx.$

Answer 1

Let $I={\int }_{-1}^1\frac{x+|x|+1}{x^2+2|x|+1}dx={\int }_{-1}^1\frac{x}{x^2+2|x|+1}dx+{\int }_{-1}^1\frac{|x|+1}{x^2+2|x|+1}dxLetf\left(x\right)=\frac{x}{x^2+2|x|+1},g\left(x\right)=\frac{|x|+1}{x^2+2|x|+1}$
$\Rightarrow f(-x)=\frac{-x}{(-x{)}^2+2|-x|+1},g(-x)=\frac{|-x|+1}{(-x{)}^2+2|-x|+1}$
$\Rightarrow f(-x)=\frac{-x}{x^2+2|x|+1},g(-x)=\frac{|x|+1}{x^2+2|x|+1}$
$\Rightarrow f(-x)=-f\left(x\right),g(-x)=g\left(x\right)\therefore f\left(x\right)$ is odd and g(x) is even function.
Therefore, $I=0+2{\int }_0^1\frac{|x|+1}{x^2+2|x|+1}dx=2{\int }_0^1\frac{x+1}{x^2+2x+1}dx=2{\int }_0^1\frac{x+1}{(x+1{)}^2}dx=2{\int }_0^1\frac{1}{x+1}dx$
$\Rightarrow I=2[log|x+1{]}_0^1=2[log2-log1]\therefore I=2log2(\because log1=0)$