What's the integral of sin(x)/x^2?

The integral of sin(x)/x^2 is -cos(x)/x + integral of cos(x)/x dx.

To solve this integral, we can use integration by parts. Let u = sin(x) and dv = 1/x^2 dx. Then du/dx = cos(x) and v = -1/x. Using the formula for integration by parts, we have:

∫ sin(x)/x^2 dx = -cos(x)/x + ∫ cos(x)/x dx

Now we need to evaluate the integral ∫ cos(x)/x dx. We can do this using integration by parts again. Let u = cos(x) and dv = 1/x dx. Then du/dx = -sin(x) and v = ln|x|. Using the formula for integration by parts, we have:

∫ cos(x)/x dx = cos(x) ln|x| + ∫ sin(x)/x dx

Substituting this back into the original integral, we get:

∫ sin(x)/x^2 dx = -cos(x)/x + cos(x) ln|x| + ∫ sin(x)/x dx

We can simplify this by adding the constant of integration to the ln term:

∫ sin(x)/x^2 dx = -cos(x)/x + cos(x) ln|x| + C

Therefore, the integral of sin(x)/x^2 is -cos(x)/x + cos(x) ln|x| + C.