The integral $\int\limits_{-1 / 2}^{1 / 2}\left\{[x]+\ln \left(\frac{1+x}{1-x}\right)\right\} d x$ equals
Answer & explanation
Correct answer: option 1
We have,
$I=\int\limits_{-1 / 2}^{1 / 2}\left\{[x]+\ln \left(\frac{1+x}{1-x}\right)\right\} d x$
$\Rightarrow I=\int\limits_{-1 / 2}^{1 / 2}[x] d x+\int\limits_{-1 / 2}^{1 / 2} \ln \left(\frac{1+x}{1-x}\right) d x$
$\Rightarrow I=\int\limits_{1 / 2}^0-1 d x+\int\limits_0^{1 / 2} 0 d x+0$ [∵ $\ln \left(\frac{1+x}{1-x}\right)$ is an odd function]
$\Rightarrow I=-1\left(0+\frac{1}{2}\right)=-\frac{1}{2}$