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

The integral of ln(x)/x^2 is -1/x + ln(x)/x + C.

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

∫ ln(x)/x^2 dx = -ln(x)/x + ∫ 1/x^2 dx

The second integral can be solved using the power rule of integration:

∫ 1/x^2 dx = -1/x + C

Substituting this back into the first equation, we get:

∫ ln(x)/x^2 dx = -ln(x)/x - 1/x + C

Simplifying this expression, we get:

∫ ln(x)/x^2 dx = -1/x + ln(x)/x + C

Therefore, the integral of ln(x)/x^2 is -1/x + ln(x)/x + C.