integral calculus problem involving integration If $2f(x)+f(-x)=(1/x)\sin(x-(1/x))$. Then what is the value of: $$\int_{e^{-1}}^{e} f (x) \mathrm {d}x $$
 A: $$2f(x)+f(-x)=\left(\frac{1}{x}\right)\sin\left(x-\frac{1}{x}\right)\\\text{Change $x$ to $-x$}\\\begin{align}2f(-x)+f(+x)&=\left(-\frac1x\right)\sin\left(-x+\frac{1}{x}\right)\\&=-\frac1x\cdot\sin\left(-\left(x-\frac1x\right)\right)\\&=\frac1x\cdot\sin\left(x-\frac1x\right)\end{align}$$
Solve this system of equations to find $f(x)$.
$$\begin{cases}2f(x)+f(-x)=\dfrac1x\sin\left(x-\dfrac1x\right)\\f(x)+2f(-x)=\dfrac1x\sin\left(x-\dfrac1x\right)\end{cases}\\\to f(x)=\dfrac13\cdot\frac1x\cdot\sin\left(x-\dfrac1x\right)$$
Now, we can solve the integral.
$$u=\frac1x \to dx=-\frac{1}{x^2} ,\\x=e=\frac1u \to u=e^{-1} \\x=e^{-1}=\frac1u \to u=e\\ \to \\
A=\int_{e^{-1}}^{e} \dfrac13\cdot\frac1x\cdot\sin\left(x-\dfrac1x\right)=\\
\int_{e}^{e^{-1}} \dfrac13\cdot u\cdot \sin\left(\frac1u-u\right)\left(-\dfrac{1}{u^2}du\right)=\\
\int_{e}^{e^{-1}} \dfrac13\cdot\frac1u\cdot\sin\left(u-\frac1u\right)du=\\
-\int_{e^{-1}}^{e} \dfrac13\cdot\frac1u\cdot\sin\left(u-\frac1u\right)du=\color{red} {-A} \\\to -A=A \to 2A=0\to A=0$$
A: *

*we have $f(-x)=f(x)$.

*with the substition $t=\frac{1}{x}$ show that $\int_{e^{-1}}^{e} (1/x) \sin(x-(1/x)) dx=0$

*Use 1. and 2. to compute $\int_{e^{-1}}^{e} f(x) dx$
