To solve the integral $\int (\ln x)^3 \, dx$, we can use integration by parts. The goal is to reduce the power of $(\ln x)^3$ progressively by choosing appropriate parts.

Solution:

Let: $u = (\ln x)^3 \quad \text{and} \quad dv = dx$ Then: $du = \frac{3(\ln x)^2}{x} \, dx \quad \text{and} \quad v = x$

Applying integration by parts: $\int (\ln x)^3 \, dx = x (\ln x)^3 - \int x \cdot \frac{3(\ln x)^2}{x} \, dx$ Simplify the integral: $= x (\ln x)^3 - 3 \int (\ln x)^2 \, dx$

Now we need to solve $\int (\ln x)^2 \, dx$, which requires another application of integration by parts.

Let: $u = (\ln x)^2 \quad \text{and} \quad dv = dx$ Then: $du = \frac{2 \ln x}{x} \, dx \quad \text{and} \quad v = x$

Applying integration by parts again: $\int (\ln x)^2 \, dx = x (\ln x)^2 - \int x \cdot \frac{2 \ln x}{x} \, dx$ Simplify this to: $= x (\ln x)^2 - 2 \int \ln x \, dx$

Now, solve $\int \ln x \, dx$ by integration by parts one last time.

Let: $u = \ln x \quad \text{and} \quad dv = dx$ Then: $du = \frac{1}{x} \, dx \quad \text{and} \quad v = x$

Thus: $\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx$ $= x \ln x - \int 1 \, dx$ $= x \ln x - x$

Now we can substitute back up the chain:

1. $\int \ln x \, dx = x \ln x - x$
2. $\int (\ln x)^2 \, dx = x (\ln x)^2 - 2 (x \ln x - x)$
3. $\int (\ln x)^3 \, dx = x (\ln x)^3 - 3 \left( x (\ln x)^2 - 2 (x \ln x - x) \right)$

Simplifying this will give the final answer. Would you like me to carry out the full simplification?

Follow-up Questions

1. Would you like to explore integration by parts in more depth?
2. Do you want a breakdown of each substitution step?
3. Are you interested in seeing alternative integration techniques for logarithmic functions?
4. Would you like a review of integration techniques involving powers of logarithmic functions?
5. Do you have similar integrals you'd like to practice with?

Tip:

For integrals involving powers of $\ln(x)$, using reduction formulas can streamline the process significantly.