How to calculate $\int_0^1 \frac{\sin(x)-x}{x^2} dx$? I would like the others to tell me how to calculate the integral $$\int_0^1 \frac{\sin(x)-x}{x^2} dx.$$

You can't solve this using elementary functions, as this integral involves the Cosine Integral.
If you're looking for a numeric result, you can use the Taylor expansion:
$$\frac{\sin(x)-x}{x^2} \sim-\frac{x}{6}+\frac{x^3}{120}-\frac{x^5}{5040}+O(x^7)$$
If you want the exact form, note that using integration by parts:
$$\int\frac{\sin(x)}{x^2}dx = -\frac{\sin(x)}{x} +\int\frac{\cos(x)}{x}dx = -\frac{\sin(x)}{x} +Ci(x)$$
So that:
$$\int_0^1\frac{\sin(x)-x}{x^2}dx = \left[Ci(x)-\log (x)-\frac{\sin (x)}{x}\right]^1_0 = Ci(1) + 1 - \gamma -\sin (1)$$
Using the fact that $\lim_{x\to 0} Ci(x) -\log(x) = \gamma$.
