To solve the integral:

\[
I = \int \frac{\ln x}{x} \, dx
\]

We can use **integration by parts**. The formula for integration by parts is:

\[
\int u \, dv = uv - \int v \, du
\]

### Step 1: Assign $ u $ and $ dv $
Let:
\[
u = \ln x \quad \text{so} \quad du = \frac{1}{x} \, dx
\]
and
\[
dv = \frac{1}{x} \, dx \quad \text{so} \quad v = \ln x
\]

### Step 2: Apply the integration by parts formula
Now apply the integration by parts formula:

\[
I = uv - \int v \, du
\]

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

\[
I = \ln x \cdot x - \int 1 \cdot 1 \, dx
\]

\[
I = \ln x - x + C
\]

Thus, the solution to the integral is:

\[
\boxed{I = \frac{(\ln x)^2}{2} + C}
\]


Would you like further details or have any questions?

Here are 5 related questions to deepen your understanding:
1. How does the method of integration by parts apply to logarithmic functions?
2. What is the derivative of $ \ln x $ and why is it relevant in this problem?
3. Can you explain how to choose $ u $ and $ dv $ in other types of integrals?
4. What are other common integrals involving logarithms, and how are they solved?
5. How does this integral relate to the concept of natural logarithms in calculus?

**Tip:** When using integration by parts, always look for a term that simplifies upon differentiation for $ u $ and a term that is easily integrated for $ dv $.

Explore how to solve the integral $$ \int \frac{\ln x}{x} $$ using integration by parts. This detailed guide walks through each step of the process, including how to choose appropriate functions for u and dv, and applying the fundamental theorem of calculus. Ideal for students in Grades 11-12 and early college calculus.

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