Question

The value of ${\int }_{-1}^1\frac{x^3+\left|x\right|+1}{x^2+2\left|x\right|+1}\mathrm{dx}$ is equal to

• A. log 2
• B. 2 log 2
• C. 2 log 3
• D. log 3

Answer 1

The correct option is B 2 log 2

$\mathrm{Let}I={\int }_{-1}^1\frac{x^3+\left|x\right|+1}{x^2+2\left|x\right|+1}\mathrm{dx}\Rightarrow I={\int }_{-1}^1\frac{x^3}{x^2+2\left|x\right|+1}\mathrm{dx}+{\int }_{-1}^1\frac{\left|x\right|+1}{x^2+2\left|x\right|+1}\mathrm{dx}\Rightarrow I={\int }_{-1}^1\frac{x^3}{{\left(\left|x\right|+1\right)}^2}\mathrm{dx}+{\int }_{-1}^1\frac{\left|x\right|+1}{{\left(\left|x\right|+1\right)}^2}\mathrm{dx}\mathrm{Now},\frac{x^3}{{\left(\left|x\right|+1\right)}^2}\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function}\mathrm{and},\frac{\left|x\right|+1}{{\left(\left|x\right|+1\right)}^2}\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{function}\therefore {\int }_{-1}^1\frac{x^3}{x^2+2\left|x\right|+1}\mathrm{dx}=0\mathrm{and},{\int }_{-1}^1\frac{\left|x\right|+1}{{\left(\left|x\right|+1\right)}^2}\mathrm{dx}=2{\int }_0^1\frac{\left|x\right|+1}{{\left(\left|x\right|+1\right)}^2}\mathrm{dx}\therefore I=2{\int }_0^1\frac{\left|x\right|+1}{{\left(\left|x\right|+1\right)}^2}\mathrm{dx}\Rightarrow I=2{\int }_0^1\frac{1}{\left|x\right|+1}\mathrm{dx}=2{\int }_0^1\frac{1}{x+1}\mathrm{dx}\Rightarrow I=2{\left(\log \left(x+1\right)\right]}_0^1=2\left(\log 2-\log 1\right)\Rightarrow I=2\log 2$