Definite Integral in functional relation format If  $$2f(x) + f(-x) = \frac{1}{x}\sin\Biggl(x-\frac{1}{x}\Biggl)$$ then find the value of 
$$\int_{\frac{1}{e}}^ef(x)dx$$(its not given that the function is odd or even)
I don't know where to start on this one I'm not much experienced in solving functional relations like these and I'm not even able to come up with an approach to solve it can someone help
 A: $\textbf{Hint:}$ $2f(x) + f(-x)=2f(-x) + f(x)\implies f(-x)=f(x)$. So $$f(x)=\dfrac{1}{3x}\sin\left(x-\dfrac{1}{x}\right)$$
$$f\left(\dfrac1x\right)=-x^2f(x)  \label{a}\tag{1}$$
$$\displaystyle\int_\frac1e^ef(x)dx=\int_\frac1e^1f(x)dx\:+\:\int_1^ef(x)dx$$ 
For $1^{st}$ integral, $y=\dfrac1x \implies \dfrac{-dy}{y^2}=dx$ with limits $\dfrac1e=x=\dfrac1y\implies y=e$ 
to $1=x=\dfrac1y\implies y=1$.
$$\implies \int_\frac1e^1f(x)dx=\displaystyle\int_e^1 f\left(\dfrac1y\right)\dfrac{-dy}{y^2}=\int_1^e f\left(\dfrac1y\right)\dfrac{dy}{y^2}\overset{(1)}{=}\int_1^e -y^2f(y)\dfrac{dy}{y^2}=-\int_1^e f(y)dy$$
Hence $$\displaystyle\int_\frac1e^ef(x)dx=0$$
A: The integral is zero.
Proof:
With $g(x) = \frac{1}{x}\sin(x-\frac{1}{x})$ we have $2f(x)+f(-x) = g(x) = g(-x) = 2f(-x) + f(x)\implies f(-x)=f(x)$ which gives
$$f(x) = \frac{1}{3 x}\sin(x-\frac{1}{x})$$
Now the integral can be split up as
$$i = \int_{\frac{1}{e}}^ef(x)\,dx =\int_{\frac{1}{e}}^1f(x)\,dx+\int_1^ef(x)\,dx=i_1 + i_2$$ 
Substituting $x\to \frac{1}{y}$ i.e. $dx \to - \frac{1}{y^2}dy$ in the second integral gives 
$$i_2=\int_1^\frac{1}{e}f(x)\,dx =  \int_{\frac{1}{e}}^1 \frac{1}{y^2}f(\frac{1}{y})\,dy=\int_{\frac{1}{e}}^1 \frac{1}{y^2}\frac{1}{3\frac{1}{y}}\sin(\frac{1}{y}-y)\,dy\\=\int_{\frac{1}{e}}^1 \frac{1}{3 y}\sin(\frac{1}{y}-y)\,dy=-\int_{\frac{1}{e}}^1 \frac{1}{3 y}\sin(y-\frac{1}{y})\,dy=- \int_{\frac{1}{e}}^1f(y)\,dy= - i_1 $$
Hence the integral in question is $i = i_1 +(- i_1) = 0$
Q.E.D.
Remark: from the proof it is clear that "$e$" can be any positive real number.
