Question

The integral ${\int }_{-1/2}^{1/2}(\left[x\right]+\log \frac{1+x}{1-x})dx$ equals

• A. $-1/2$
• B. $0$
• C. $1$
• D. $-2\log 2$

Answer 1

Answer: (A)

$-1/2$

$I={\int }_{-\frac{1}{2}}^{\frac{1}{2}}\left(\right[x]+log\frac{1+x}{1-x}).dx$

$I={\int }_{-\frac{1}{2}}^{\frac{1}{2}}(-1-[x]+log\frac{1-x}{1+x}).dx$

Hence

$2I={\int }_{\frac{-1}{2}}^{\frac{1}{2}}-1.dx$

$2I=-2{\int }_0^{\frac{1}{2}}1.dx$

$I=-{\int }_0^{\frac{1}{2}}1.dx$

$=-\frac{1}{2}$