Practicing Success
The integral $\int\limits_{-1 / 2}^{1 / 2}\left\{[x]+\ln \left(\frac{1+x}{1-x}\right)\right\} d x$ equals |
$-\frac{1}{2}$ 0 1 $2 \ln \left(\frac{1}{2}\right)$ |
$-\frac{1}{2}$ |
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] $\Rightarrow I=-1\left(0+\frac{1}{2}\right)=-\frac{1}{2}$ |