The integral given in the image is:

$\int \frac{\ln x}{x^2} \, dx$

To solve this integral, we can use the method of integration by parts. Recall the formula for integration by parts:

$\int u \, dv = uv - \int v \, du$

Let $u = \ln x$, so that $du = \frac{1}{x} \, dx$.
Let $dv = \frac{1}{x^2} \, dx$, so that $v = -\frac{1}{x}$.

Now apply the formula:

$\int \frac{\ln x}{x^2} \, dx = u v - \int v \, du$

Substitute $u = \ln x$, $v = -\frac{1}{x}$, and $du = \frac{1}{x} \, dx$:

$\int \frac{\ln x}{x^2} \, dx = \left( \ln x \cdot -\frac{1}{x} \right) - \int \left( -\frac{1}{x} \cdot \frac{1}{x} \right) dx$

Simplify:

$\int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx$

The second integral is straightforward:

$\int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = -x^{-1} = -\frac{1}{x}$

Now substitute back:

$\int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} - \frac{1}{x} + C$

Final answer:

$\int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} - \frac{1}{x} + C$

Let me know if you'd like further clarification or details on this solution.

Here are 5 related questions for further exploration:

1. How does integration by parts work for different types of functions?
2. What are some practical applications of logarithmic integrals in physics or engineering?
3. How can substitution simplify other integrals with logarithmic terms?
4. How does the constant of integration $C$ influence the solution?
5. What happens if we change the power of $x$ in the denominator?

Tip: When choosing $u$ in integration by parts, pick the function that simplifies upon differentiation.