Question

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

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

Answer 1

The correct option is B $x{e}^{x+\frac{1}{x}}+c$

$I=\int (1+x-\frac{1}{x})e^{x+\frac{1}{x}}dx$
$=\int [e^{x+\frac{1}{x}}+x.e^{x+\frac{1}{x}}(1-\frac{1}{x^2})]dx$
Since, $\int \left(f\right(x)+x{f}^{\prime }(x\left)\right)dx=xf\left(x\right)+c$
Thus, $I=x{e}^{x+\frac{1}{x}}+c$