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Question Number 136279 by aupo14 last updated on 20/Mar/21

Answered by mindispower last updated on 20/Mar/21

=∫e^(x+(1/x)) dx+∫x(1−(1/x^2 ))e^(x+(1/x)) dx=A 2nd by part ∫x(1−(1/x^2 ))e^(x+(1/x)) dx=xe^(x+(1/x)) −∫e^(x+(1/x)) ⇔∫x(1−(1/x^2 ))e^(x+(1/x)) dx+∫e^(x+(1/x)) dx=xe^(x+(1/x)) ⇔∫(1+x−(1/x))e^(x+(1/x)) dx=xe^(x+(1/x)) +c

$$=\int{e}^{{x}+\frac{\mathrm{1}}{{x}}} {dx}+\int{x}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){e}^{{x}+\frac{\mathrm{1}}{{x}}} {dx}={A} \\ $$$$\mathrm{2}{nd}\:{by}\:{part} \\ $$$$\int{x}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){e}^{{x}+\frac{\mathrm{1}}{{x}}} {dx}={xe}^{{x}+\frac{\mathrm{1}}{{x}}} −\int{e}^{{x}+\frac{\mathrm{1}}{{x}}} \\ $$$$\Leftrightarrow\int{x}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){e}^{{x}+\frac{\mathrm{1}}{{x}}} {dx}+\int{e}^{{x}+\frac{\mathrm{1}}{{x}}} {dx}={xe}^{{x}+\frac{\mathrm{1}}{{x}}} \\ $$$$\Leftrightarrow\int\left(\mathrm{1}+{x}−\frac{\mathrm{1}}{{x}}\right){e}^{{x}+\frac{\mathrm{1}}{{x}}} {dx}={xe}^{{x}+\frac{\mathrm{1}}{{x}}} +{c} \\ $$

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