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)

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