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

Answered by mindispower last updated on 20/Mar/21

$$=\int e^{x+\frac{1}{x}}dx+\int x(1-\frac{1}{x^2})e^{x+\frac{1}{x}}dx=A$$$$2ndbypart$$$$\int x(1-\frac{1}{x^2})e^{x+\frac{1}{x}}dx={xe}^{x+\frac{1}{x}}-\int e^{x+\frac{1}{x}}$$$$\iff \int x(1-\frac{1}{x^2})e^{x+\frac{1}{x}}dx+\int e^{x+\frac{1}{x}}dx={xe}^{x+\frac{1}{x}}$$$$\iff \int (1+x-\frac{1}{x})e^{x+\frac{1}{x}}dx={xe}^{x+\frac{1}{x}}+c$$

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