Question

Solve :
$\int (1+x-x^{-1})e^{x+x^{-1}}dx=$

Answer 1

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

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

substitute $u=x+\frac{1}{x}\to du=(1-\frac{1}{x^2})dx$

$\int [1+x(1-\frac{1}{x^2})]e^{u}dx$

$=\int [e^{u}dx+x{e}^{u}du]$

$=\int \left[d\right(x{e}^u\left)\right]$

$=x{e}^u$

$=x{e}^{x+\frac{1}{x}}+C$