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

Question

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

I was trying to convert it in the form:

$\int e^{h(x)}(h'(x).f(x)+f'(x))=e^{h(x)}f(x)+C$ but it does reduce to that form . Could someone suggest an alternate approach?

Answer 1

Doesn't $h(x) = x+\frac{1}{x}, f(x) = x$ work?

Answer 2

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

$=\int e^{x+\frac{1}{x}}\left(x-\dfrac{1}{x}\right)dx+\int e^{x+\frac{1}{x}}~dx$

$=\int xe^{x+\frac{1}{x}}\left(1-\dfrac{1}{x^2}\right)dx+\int e^{x+\frac{1}{x}}~dx$

$=\int x~d\left(e^{x+\frac{1}{x}}\right)+\int e^{x+\frac{1}{x}}~dx$

$=xe^{x+\frac{1}{x}}-\int e^{x+\frac{1}{x}}~dx+\int e^{x+\frac{1}{x}}~dx$

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