Let's analyze the integral presented in the image.

Problem:

You are given the following integral:

$\int \frac{e^{2z} \ln((e^{2z} + 6)^5)}{(e^{2z} + 6)^3} \, dz$

The instruction suggests using a substitution $x = e^{2z} + 6$, and then applying integration by parts with $u = \ln(x^5)$.

Step-by-Step Approach:

1. Substitution: Set $x = e^{2z} + 6$. The differential $dx$ can be found by differentiating $x$ with respect to $z$:

   $dx = 2e^{2z} dz \quad \Rightarrow \quad dz = \frac{dx}{2e^{2z}}$

   From the substitution $x = e^{2z} + 6$, we also have:

   $e^{2z} = x - 6$

   So, $dz = \frac{dx}{2(x - 6)}$.

2. Simplifying the Integral: After substituting, the given integral becomes:

   $\int \frac{(x - 6) \ln(x^5)}{x^3} \cdot \frac{dx}{2(x - 6)}$

   Simplifying the logarithmic expression $\ln(x^5) = 5 \ln(x)$, the integral simplifies to:

   $\frac{1}{2} \int \frac{5 \ln(x)}{x^3} \, dx$

3. Applying Integration by Parts: Now, we apply integration by parts with $u = \ln(x)$ and $dv = \frac{1}{x^3} \, dx$.

   • $du = \frac{1}{x} dx$
   • $v = -\frac{1}{2x^2}$ (since $\int x^{-3} dx = -\frac{1}{2x^2}$)

   The formula for integration by parts is:

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

   Applying this to our integral:

   $\frac{1}{2} \left[ 5 \ln(x) \left( -\frac{1}{2x^2} \right) - \int \left( -\frac{1}{2x^2} \right) \cdot \frac{5}{x} dx \right]$

4. Resulting Functions $F(x)$ and $G(x)$: After simplifying, we obtain:

   • $F(x) = -\frac{5 \ln(x)}{4x^2}$
   • $G(x) = \frac{5}{4x^3}$

Final Answer:

• $F(x)$: $-\frac{5 \ln(x)}{4x^2}$
• $G(x)$: $\frac{5}{4x^3}$

Would you like more details on any of these steps or a further breakdown?

Here are five related questions:

1. How is integration by parts derived?
2. What other substitution techniques can simplify logarithmic integrals?
3. How does changing the base of the logarithm affect integration problems?
4. What are some alternative methods for solving integrals involving $e^x$ terms?
5. How does the chain rule apply in substitutions for integrals?

Tip: Integration by parts is particularly useful when you have a product of functions where one function becomes simpler upon differentiation.