Math Problem Statement

Consider the following integral: ∫ (e^(-2z) * ln((e^(-2z) + 3)^4) / (e^(-2z) + 3)^3) dz. If the substitution x = e^(-2z) + 3 is made and integration by parts is applied using u = ln(x^4), the integral can be written as F(x) + ∫ G(x) dx. Identify the functions F(x) and G(x).

Solution

The image presents an integral problem with the following expression:

e2zln((e2z+3)4)(e2z+3)3dz\int \frac{e^{-2z} \ln\left( (e^{-2z} + 3)^4 \right)}{(e^{-2z} + 3)^3} \, dz

The problem suggests making the substitution:

x=e2z+3x = e^{-2z} + 3

and using integration by parts with:

u=ln(x4)u = \ln(x^4)

This results in an expression of the form:

F(x)+G(x)dxF(x) + \int G(x) \, dx

The task is to identify the functions F(x)F(x) and G(x)G(x).

Let’s break it down step by step:

  1. Substitution:

    • x=e2z+3x = e^{-2z} + 3
    • This gives dz=12dxx3dz = -\frac{1}{2} \frac{dx}{x - 3}

    After substituting into the integral, we transform it into a new variable xx.

  2. Integration by Parts:

    • The choice of u=ln(x4)=4ln(x)u = \ln(x^4) = 4 \ln(x) and the differentiation will lead to the splitting of the integral into F(x)F(x) and G(x)G(x).

It seems like you're given options for the functions F(x)F(x) and G(x)G(x). Without seeing those choices, I can't provide the exact forms, but based on the setup, I would expect:

  • F(x)F(x) to be a part derived from the logarithmic term after integration by parts.
  • G(x)G(x) would involve the remaining part of the integrand after applying the product rule during integration by parts.

Do you want to go into more detail on how to solve this or clarify any steps?


Here are five questions you might explore further:

  1. What is the derivative of u=ln(x4)u = \ln(x^4), and how does it affect the integration by parts setup?
  2. How do you handle the integral involving ln(x)\ln(x)?
  3. What role does the substitution x=e2z+3x = e^{-2z} + 3 play in simplifying the original integral?
  4. What are typical strategies for applying integration by parts in logarithmic integrals?
  5. Can you identify patterns in integrals involving exponential and logarithmic functions?

Tip: When using integration by parts, always remember to carefully choose uu and dvdv to simplify the resulting integrals.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Integration by Parts
Logarithmic Integration
Substitution

Formulas

Integration by parts: ∫ u dv = uv - ∫ v du
Substitution: x = e^(-2z) + 3

Theorems

Integration by Parts Theorem

Suitable Grade Level

University-level Calculus