Discussion Forum

 The integral  $\int_{}^{} (1+x-\frac{1}{x})e^{x+\frac{1}{x}}dx$  id equal to 

Answer:

Explanation:

$\int_{}^{} (1+x-\frac{1}{x})e^{x+\frac{1}{x}}dx$

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

 =  $\int_{}^{} e^{x+1/x}dx+xe^{x+1/x}-\int_{}^{} \frac{d}{dx}(x)e^{x+1/x} dx$

    =  $\int_{}^{} e^{x+1/x}dx+xe^{x+1/x}-\int_{}^{} e^{x+\frac{1}{x}}dx$

   $\left[\because \int_{}^{} \left(1-\frac{1}{x^{2}}\right)e^{x+1/x}dx=e^{x+1/x}\right]$

    $=\int_{}^{} e^{x+1/x}dx+xe^{x+1/x}-\int_{}^{} ex^{+1/x}dx$

  =  $xe^{x+1/x}+C$