Question

The value of $\int e^{(x+\frac{1}{x})}(x-\frac{1}{x}+1)dx$ is

• A. $(x+\frac{1}{x})+C$
• B. $(x-\frac{1}{x}+1)e^{(x+\frac{1}{x})}+C$
• C. $(x+1)e^{(x+\frac{1}{x})}+C$
• D. $x{e}^{(x+\frac{1}{x})}+C$

Answer 1

The correct option is D $x{e}^{(x+\frac{1}{x})}+C$

$\int e^{(x+\frac{1}{x})}(x-\frac{1}{x}+1)dx=\int e^{(x+\frac{1}{x})}\left(1\right)dx+\int e^{(x+\frac{1}{x})}(x-\frac{1}{x})dx=x{e}^{(x+\frac{1}{x})}-\int x{e}^{(x+\frac{1}{x})}\times (1-\frac{1}{x^2})dx+\int e^{(x+\frac{1}{x})}(x-\frac{1}{x})dx=x{e}^{(x+\frac{1}{x})}+C$