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Question Number 101650 by I want to learn more last updated on 03/Jul/20

Commented by mr W last updated on 03/Jul/20

$$f\left(x\right)=a^{x}fora\in R^{+}$$

Answered by mathmax by abdo last updated on 03/Jul/20

$$\mathrm{I}={\int }_{-1}^1\frac{\mathrm{dx}}{1+\mathrm{f}\left(\mathrm{x}\right)}\mathrm{cha}7\mathrm{gement}\mathrm{x}=-\mathrm{t}\mathrm{give}\mathrm{I}=-{\int }_{-1}^1\frac{-\mathrm{dt}}{1+\mathrm{f}(-\mathrm{t})}$$$$={\int }_{-1}^1\frac{\mathrm{dt}}{1+\frac{1}{\mathrm{f}\left(\mathrm{t}\right)}}={\int }_{-1}^1\frac{\mathrm{f}\left(\mathrm{t}\right)}{1+\mathrm{f}\left(\mathrm{t}\right)}\mathrm{dt}\Rightarrow 2\mathrm{I}={\int }_{-1}^1\frac{\mathrm{dx}}{1+\mathrm{f}\left(\mathrm{x}\right)}+{\int }_{-1}^1\frac{\mathrm{f}\left(\mathrm{x}\right)}{1+\mathrm{f}\left(\mathrm{x}\right)}\mathrm{dx}$$$$={\int }_{-1}^1\mathrm{dx}=2\Rightarrow \mathrm{I}=1$$

Commented by I want to learn more last updated on 04/Jul/20

$$\mathrm{Thanks}\mathrm{sir}$$

Commented by mathmax by abdo last updated on 04/Jul/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Answered by MAB last updated on 03/Jul/20

$${\int }_{-1}^0\frac{1}{1+f\left(x\right)}dx\stackrel{u=-x}{=}{\int }_0^1\frac{1}{1+f(-u)}du$$$$={\int }_0^1\frac{1}{1+\frac{1}{f\left(u\right)}}du(f(-u)=\frac{1}{f\left(u\right)})$$$$={\int }_0^1\frac{f\left(u\right)}{1+f\left(u\right)}du$$$$thus$$$${\int }_{-1}^1\frac{1}{1+f\left(x\right)}dx={\int }_0^1\frac{1}{1+f\left(x\right)}+\frac{f\left(x\right)}{1+f\left(x\right)}dx$$$$={\int }_0^{1}1dx$$$$=1$$

Commented by MAB last updated on 05/Jul/20

$$youarewelcome$$

Commented by mr W last updated on 03/Jul/20

$$verynicesir!$$$${\int }_{-1}^1\frac{dx}{1+a^x}=1foranya\in R^{+}$$

Commented by I want to learn more last updated on 04/Jul/20

$$\mathrm{Thanks}\mathrm{sir}.$$

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