$\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x$, is equal to
Answer & explanation
Correct answer: option 1
Let $I=\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x$. Then,
$I=\int e^{x+x^{-1}} d x+\int x\left(1-\frac{1}{x^2}\right) e^{x+\frac{1}{x}} d x$
$\Rightarrow I=\int e^{x+x^{-1}} d x+x e^{x+\frac{1}{x}}-\int e^{x+x^{-1}} d x+C$
$\Rightarrow I=x e^{x+x^{-1}}+C$