The integral $\int_{\frac{-1}{2}}^{\frac{1}{2}}\left([x]+\log _c\left(\frac{1+x}{1-x}\right)\right) \mathrm{d} x$, where $[x]$ represent greatest integer function, equals

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Aicharya · Accepted Answer

Let $\mathrm{I}=\int_{\frac{-1}{2}}^{\frac{1}{2}}\left([x]+\log _{\mathrm{e}}\left(\frac{1+x}{1-x}ight)ight) \mathrm{d} x$ $=\int_{\frac{-1}{2}}^0[x] \mathrm{d} x+\int_0^{\frac{1}{2}}[x] \mathrm{d} x+\int_{\frac{-1}{2}}^{\frac{1}{2}} \log _{\mathrm{e}}\left(\frac{1+x}{1-x}ight) \mathrm{d} x$ Let $g(x)=\log \left(\frac{1+x}{1-x}ight)$ $g(-x)=\log \left(\frac{1-x}{1+x}ight)=-\log \left(\frac{1+x}{1-x}ight)=-g(x)$ $	herefore \quad \mathrm{g}(x)$ is a odd function. $	herefore \quad \int_{\frac{-1}{2}}^{\frac{1}{2}} \mathrm{~g}(x) \mathrm{d} x=0$ $\begin{aligned} 	herefore \quad \mathrm{I} & =\int_{\frac{-1}{2}}^0(-1) \mathrm{d} x+\int_0^{\frac{1}{2}}(0) \mathrm{d} x+0 \ & =[-x]_{\frac{-1}{2}}^0+0 \ & =\frac{-1}{2}\end{aligned}$

The integral $\int_{\frac{-1}{2}}^{\frac{1}{2}}\left([x]+\log _c\left(\frac{1+x}{1-x}\right)\right)…

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