Finding $\int_{1/4}^4 \frac{1}{x}\sin(x-\frac{1}{x})\,dx $ Question:

Find the value of: $$\int_{1/4}^4 \frac{1}{x}\sin(x-\frac{1}{x})\,dx $$

Attempt:
I seriously have no idea how to attempt this question. I suppose I will first have to treat this as an indefinite integral to find the integration of the function, and then substitute the limits to find the value.
I tried to do this using the by-parts method, however to no avail. I can't think of what to substitute in order to solve the integral. Any hint would be appreciated.

If we substitute $t=x-\frac{1}{x}$ (we are allowed since $g(x)=x-\frac{1}{x}$ is increasing and differentiable over $\left[\frac{1}{4},4\right]$), we are left with:
$$ I = \int_{-15/4}^{15/4}\sin(t)\frac{dt}{\sqrt{4+t^2}}$$
hence we are integrating an odd integrable function over a symmetric domain with respect to the origin, so the value of the integral is simply zero.
