Question Number 86734 by Tony Lin last updated on 30/Mar/20
![Find all functions that satisfy the equation [∫f(x)dx][∫(1/(f(x)))dx]=−1](https://www.tinkutara.com/question/Q86734.png)
$${Find}\:{all}\:{functions}\:{that}\:{satisfy}\:{the} \\ $$$${equation} \\ $$$$\left[\int{f}\left({x}\right){dx}\right]\left[\int\frac{\mathrm{1}}{{f}\left({x}\right)}{dx}\right]=−\mathrm{1} \\ $$
Answered by mr W last updated on 30/Mar/20
![∫(1/(f(x)))dx=−(1/(∫f(x)dx)) (1/(f(x)))=((f(x))/((∫f(x)dx)^2 )) ⇒∫f(x)dx=±f(x) ⇒f(x)=±f′(x) ∫((df(x))/(f(x)))=±∫dx ln f(x)=±x+c ⇒f(x)=Ce^(±x) check: ∫f(x)dx=±Ce^(±x) ∫(1/(f(x)))dx=±∫(1/C)e^(∓x) dx=∓(1/C)e^(∓x) [∫f(x)dx][∫(1/(f(x)))dx]=(±Ce^(±x) )(∓(1/C)e^(±x) )=−1](https://www.tinkutara.com/question/Q86738.png)
$$\int\frac{\mathrm{1}}{{f}\left({x}\right)}{dx}=−\frac{\mathrm{1}}{\int{f}\left({x}\right){dx}} \\ $$$$\frac{\mathrm{1}}{{f}\left({x}\right)}=\frac{{f}\left({x}\right)}{\left(\int{f}\left({x}\right){dx}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\int{f}\left({x}\right){dx}=\pm{f}\left({x}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\pm{f}'\left({x}\right) \\ $$$$\int\frac{{df}\left({x}\right)}{{f}\left({x}\right)}=\pm\int{dx} \\ $$$$\mathrm{ln}\:{f}\left({x}\right)=\pm{x}+{c} \\ $$$$\Rightarrow{f}\left({x}\right)={Ce}^{\pm{x}} \\ $$$$ \\ $$$${check}: \\ $$$$\int{f}\left({x}\right){dx}=\pm{Ce}^{\pm{x}} \\ $$$$\int\frac{\mathrm{1}}{{f}\left({x}\right)}{dx}=\pm\int\frac{\mathrm{1}}{{C}}{e}^{\mp{x}} {dx}=\mp\frac{\mathrm{1}}{{C}}{e}^{\mp{x}} \\ $$$$\left[\int{f}\left({x}\right){dx}\right]\left[\int\frac{\mathrm{1}}{{f}\left({x}\right)}{dx}\right]=\left(\pm{Ce}^{\pm{x}} \right)\left(\mp\frac{\mathrm{1}}{{C}}{e}^{\pm{x}} \right)=−\mathrm{1} \\ $$
Commented by Tony Lin last updated on 30/Mar/20
$${thanks}\:{sir} \\ $$