Question

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

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

Answer 1

Answer: (D)

$I=\int\left\{e^{\left(x+\frac{1}{x}\right)}+x\left(1-\frac{1}{x^{2}}\right) e^{x+\frac{1}{x}}\right\} d x $
$=x \cdot e^{x+\frac{1}{x}}+c$
As $\int\left(x f^{\prime}(x)+f(x)\right) d x=x f(x)+c$