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Question Number 146790 by puissant last updated on 15/Jul/21

$$\int \frac{1}{\mathrm{x}}{\mathrm{e}}^{-\frac{1}{{\mathrm{x}}^2}}\mathrm{dx}$$

Answered by Cyriille last updated on 15/Jul/21

$$\mathbf{I}=\int \frac{1}{x}{e}^{-\frac{1}{x^2}}dx$$$$letu=e^{-\frac{1}{x^2}}$$$$\Rightarrow du=\frac{1}{x}{e}^{-\frac{1}{x^2}}dx$$$$\Rightarrow \mathbf{I}=\int du=u+k,k\in \mathbf{R}$$$$\Rightarrow \mathbf{I}=e^{-\frac{1}{x^2}}+k,k\in \mathbf{R}$$

Commented by puissant last updated on 15/Jul/21

$$\mathrm{at}\mathrm{the}\mathrm{level}\mathrm{of}\mathrm{the}\mathrm{third}\mathrm{line}\mathrm{what}$$$$\mathrm{is}\mathrm{the}\mathrm{derivative}\mathrm{sir}.?$$

Commented by mathmax by abdo last updated on 16/Jul/21

$$\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{-\frac{1}{{\mathrm{x}}^2}}\right)={(-\frac{1}{{\mathrm{x}}^2})}^{{}^{\prime }}{\mathrm{e}}^{-\frac{1}{{\mathrm{x}}^2}}=\frac{2\mathrm{x}}{{\mathrm{x}}^4}{\mathrm{e}}^{-\frac{1}{{\mathrm{x}}^2}}=\frac{2}{{\mathrm{x}}^3}{\mathrm{e}}^{-\frac{1}{{\mathrm{x}}^2}}$$$$\mathrm{answer}\mathrm{given}\mathrm{not}\mathrm{correct}...!$$

Commented by puissant last updated on 16/Jul/21

$$\mathrm{done}\mathrm{then}\mathrm{sir}$$

Commented by Cyriille last updated on 16/Jul/21

$$\mathrm{By}\mathrm{definition},$$$$\frac{d}{dx}\left(e^{f\left(x\right)}\right)=f^{{}^{\prime }}\left(x\right)e^{f\left(x\right)}$$$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{x^2}\right)=\frac{d}{dx}\left(x^{-2}\right)=-x^{-1}=-\frac{1}{x}$$$$\frac{d}{dx}\left(e^{-\frac{1}{x^2}}\right)=-{\left(\frac{1}{x^2}\right)}^{{}^{\prime }}{e}^{-\frac{1}{x^2}}=-(-\frac{1}{x})e^{-\frac{1}{x^2}}$$$$\Rightarrow \frac{d}{dx}\left(e^{-\frac{1}{x^2}}\right)=\frac{1}{x}{e}^{-\frac{1}{x^2}}$$$$$$

Commented by Cyriille last updated on 16/Jul/21

$$\Rightarrow \int d\left(e^{-\frac{1}{x^2}}\right)=\int \frac{1}{x}{e}^{-\frac{1}{x^2}}dx$$$$,\int \left(\frac{1}{x}{e}^{-\frac{1}{x^2}}\right)dx=e^{-\frac{1}{x^2}}$$

Commented by mathmax by abdo last updated on 16/Jul/21

$$\mathrm{d}\left({\mathrm{x}}^{-2}\right)=-{2\mathrm{x}}^{-3}=-\frac{2}{{\mathrm{x}}^3}\Rightarrow \mathrm{d}(-{\mathrm{x}}^{-2})=\frac{2}{{\mathrm{x}}^3}$$

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