Math Problem Statement

Evaluate the integral: ∫(8x / (4x^2 + 7)) dx from 2 to 5.

Solution

We are tasked with evaluating the definite integral:

258x4x2+7dx\int_{2}^{5} \frac{8x}{4x^2 + 7} \, dx

Solution Outline:

  1. Substitution Method: We'll use substitution to simplify the integrand. Let: u=4x2+7u = 4x^2 + 7 Then: dudx=8xdu=8xdx\frac{du}{dx} = 8x \quad \Rightarrow \quad du = 8x \, dx Now the integral becomes: u(2)u(5)duu\int_{u(2)}^{u(5)} \frac{du}{u} where u(2)=4(22)+7=23u(2) = 4(2^2) + 7 = 23 and u(5)=4(52)+7=107u(5) = 4(5^2) + 7 = 107.

  2. Integral of 1u\frac{1}{u}: The integral is straightforward: duu=lnu\int \frac{du}{u} = \ln |u| Hence, the definite integral becomes: ln(u)23107=ln(107)ln(23)=ln(10723)\ln(u) \Big|_{23}^{107} = \ln(107) - \ln(23) = \ln\left(\frac{107}{23}\right)

  3. Final Result: ln(10723)\ln\left(\frac{107}{23}\right)

I'll now compute the final value of the expression.

Let me calculate.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can still compute the value of:

ln(10723)\ln\left(\frac{107}{23}\right)

using a calculator or logarithm tool to get the final numerical result.

Would you like further details on this process or have any questions?

Here are five related questions that can expand on this:

  1. How does substitution work in solving integrals?
  2. What other methods are available to solve definite integrals?
  3. What is the significance of the natural logarithm in this context?
  4. Can definite integrals always be evaluated using substitution?
  5. How does changing the limits of integration work when substituting?

Tip: When solving integrals, always check if substitution or integration by parts simplifies the problem!

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Substitution Method
Definite Integrals
Logarithmic Functions

Formulas

Substitution: u = 4x^2 + 7
Definite Integral: ∫ f'(x) / f(x) dx = ln|f(x)| + C

Theorems

Fundamental Theorem of Calculus
Substitution Theorem

Suitable Grade Level

Undergraduate Calculus (Calculus I)