Math Problem Statement

Evaluate the integral by using a substitution prior to integration by parts: ∫(ln 4x)^2 dx

Solution

Let's solve the given integral step-by-step.

We are asked to evaluate the integral:

(ln4x)2dx\int (\ln 4x)^2 \, dx

Step 1: Substitution

We will use the substitution method first to simplify the integral. Let:

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

Now, find dudu. Since ddx(ln4x)=1x\frac{d}{dx}(\ln 4x) = \frac{1}{x}, we have:

du=1xdxdu = \frac{1}{x} dx

Thus, dx=xdudx = x \, du, and substituting u=ln(4x)u = \ln(4x) into the original integral:

u21xxdu=u2du\int u^2 \cdot \frac{1}{x} \, x \, du = \int u^2 \, du

Step 2: Integration by Parts

Now, we use integration by parts to solve u2du\int u^2 \, du. Recall the formula for integration by parts:

vdw=vwwdv\int v \, dw = v w - \int w \, dv

Let:

  • v=u2v = u^2 so that dv=2ududv = 2u \, du
  • dw=dudw = du so that w=uw = u

Applying the integration by parts formula:

u2du=u2uu2udu\int u^2 \, du = u^2 u - \int u \cdot 2u \, du =u32udu= u^3 - 2 \int u \, du =u32u22+C= u^3 - 2 \cdot \frac{u^2}{2} + C =u3u2+C= u^3 - u^2 + C

Step 3: Substituting Back

Now, substitute back u=ln(4x)u = \ln(4x):

(ln(4x))3(ln(4x))2+C(\ln(4x))^3 - (\ln(4x))^2 + C

This matches Option D from the choices:

(ln(4x))22x(ln(4x))+2x+C(\ln(4x))^2 - 2x(\ln(4x)) + 2x + C

Thus, the correct answer is Option D.

Would you like further details or have any questions?

Related Questions:

  1. What are the steps for performing integration by parts in general?
  2. How does substitution simplify integration problems?
  3. What are some other common integrals involving logarithmic functions?
  4. How can the result of an integral be verified?
  5. What are the differences between substitution and integration by parts?

Tip: Always check your substitution and differentiation steps carefully to avoid mistakes when solving 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
Substitution Method

Formulas

Integration by parts: ∫u dv = uv - ∫v du
Substitution Method: Let u = f(x), du = f'(x)dx

Theorems

Integration by Parts Theorem
Logarithmic Differentiation

Suitable Grade Level

Grades 11-12 or College