Question

${\int }_{\frac{1}{2}}^2(1-\frac{1}{x^2})log\left(\frac{1-x+x^2}{1+x+x^2}\right)dx$ is equal to

• A. 2 log 2
• B. log 2
• C. log 3

Answer 1

$I={\int }_{\frac{1}{2}}^2(1-\frac{1}{x^2})log\left(\frac{1-x+x^2}{1+x+x^2}\right)dx$
Put $x=\frac{1}{y}\Rightarrow dx=-\frac{1}{y^2}dy\therefore I={\int }_2^{\frac{1}{2}}(1-y^2)log\left(\frac{1-\frac{1}{y}+\frac{1}{y^2}}{1+\frac{1}{y}+\frac{1}{y^2}}\right)(-\frac{dy}{y^2})=-{\int }_{\frac{1}{2}}^2(1-y^2)log\left(\frac{y^2-y+1}{y^2+y+1}\right)\left(\frac{dy}{y^2}\right)={\int }_{\frac{1}{2}}^2(\frac{1}{y^2}-1)log\left(\frac{1-y+y^2}{1+y+y^2}\right)dy=-I\Rightarrow I=0$