Mathematics

The integral $\displaystyle \int (1+x-\displaystyle \frac{1}{x})e^{x+\frac{1}{x}} dx$ is equal to

$xe^{x+\frac{1}{x}} +c$

SOLUTION
$\displaystyle \int (1+x-\displaystyle \frac{1}{x})e^{x+\frac{1}{x}} dx$

$\displaystyle =\int e^{(x+\frac{1}{x})}dx + \int x(1-\displaystyle \frac{1}{x^{2}})e^{(x+\frac{1}{x})}dx$

$\displaystyle =\int e^{(x+\frac{1}{x})}dx + xe^{(x+\frac{1}{x})}-\int e^{(x+\frac{1}{x})}dx$

Using integration by parts

$=xe^{(x+\frac{1}{x})}+c$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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