Question

If $\int (1+x-\frac{1}{x})e^{x+\frac{1}{x}}dx=f\left(x\right)e^{x+\frac{1}{x}}+c$, where c is a constant and x $\ne$ 0, then f(2017) =

• A. 2016
• B. 2017
• C. 4034
• D. 4032

Answer 1

The correct option is B 2017

$\int (1+x-\frac{1}{x})e^{x+\frac{1}{x}}dx$
$=\int [1+x(1-\frac{1}{x^2})]e^{x+\frac{1}{x}}dx$
$=\int e^{x+\frac{1}{x}}dx+\int x(1-\frac{1}{x^2})e^{x+\frac{1}{x}}dx,$ applying by parts for second integral
$=\int e^{x+\frac{1}{x}}dx+x{e}^{x+\frac{1}{x}}-\int e^{x+\frac{1}{x}}dx$
$=x{e}^{x+\frac{1}{x}}+C=f\left(x\right)e^{x+\frac{1}{x}}+C$
$\Rightarrow f\left(x\right)=x\therefore f\left(2017\right)=2017$