\( \int _ { 1 } \frac { \ln x } { x ^ { 2 } } d x \)

To evaluate the integral: \[ \int \frac{\ln x}{x^2} \, dx \] we can use **integration by parts**. Recall the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] **Step 1: Choose \( u \) and \( dv \)** Let: - \( u = \ln x \) ⇒ \( du = \frac{1}{x} \, dx \) - \( dv = \frac{1}{x^2} \, dx \) ⇒ \( v = -\frac{1}{x} \) **Step 2: Apply Integration by Parts** Plugging into the formula: \[ \int \frac{\ln x}{x^2} \, dx = uv - \int v \, du \] \[ = \ln x \left(-\frac{1}{x}\right) - \int \left(-\frac{1}{x}\right) \left(\frac{1}{x}\right) \, dx \] \[ = -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx \] **Step 3: Integrate the Remaining Integral** \[ \int \frac{1}{x^2} \, dx = -\frac{1}{x} + C \] **Step 4: Combine the Results** \[ -\frac{\ln x}{x} - \frac{1}{x} + C = -\frac{\ln x + 1}{x} + C \] **Final Answer:** \[ \int \frac{\ln x}{x^2} \, dx = -\frac{\ln x + 1}{x} + C \] where \( C \) is the constant of integration.