Math Problem Statement

If ∫(2x + 3)lnx dx = yz - ∫z dy, then yz = ....

Solution

We are given the equation:

(2x+3)lnxdx=yzzdy\int (2x + 3) \ln x \, dx = y z - \int z \, dy

The objective is to find the expression for yzyz and match it with one of the given options.

Let's break down the process:

Step 1: Solve the Integral

We need to evaluate:

(2x+3)lnxdx\int (2x + 3) \ln x \, dx

This can be split into two parts using the distributive property of integrals:

(2x+3)lnxdx=2xlnxdx+3lnxdx\int (2x + 3) \ln x \, dx = \int 2x \ln x \, dx + \int 3 \ln x \, dx

Now, we handle each integral separately:

  1. For 2xlnxdx\int 2x \ln x \, dx:
    This is an integration by parts problem. Let:

    • u=lnxu = \ln x, so du=1xdxdu = \frac{1}{x} dx
    • dv=2xdxdv = 2x \, dx, so v=x2v = x^2

    Using integration by parts formula udv=uvvdu\int u \, dv = uv - \int v \, du, we get: 2xlnxdx=x2lnxx21xdx=x2lnxxdx\int 2x \ln x \, dx = x^2 \ln x - \int x^2 \cdot \frac{1}{x} dx = x^2 \ln x - \int x \, dx =x2lnxx22= x^2 \ln x - \frac{x^2}{2}

  2. For 3lnxdx\int 3 \ln x \, dx:
    Again, using integration by parts, let:

    • u=lnxu = \ln x, so du=1xdxdu = \frac{1}{x} dx
    • dv=3dxdv = 3 \, dx, so v=3xv = 3x

    Applying the formula, we get: 3lnxdx=3xlnx3x1xdx=3xlnx3dx\int 3 \ln x \, dx = 3x \ln x - \int 3x \cdot \frac{1}{x} dx = 3x \ln x - \int 3 \, dx =3xlnx3x= 3x \ln x - 3x

Step 2: Combine the Results

Now, combine both integrals:

(2x+3)lnxdx=(x2lnxx22)+(3xlnx3x)\int (2x + 3) \ln x \, dx = \left( x^2 \ln x - \frac{x^2}{2} \right) + (3x \ln x - 3x)

Simplifying:

=x2lnx+3xlnxx223x= x^2 \ln x + 3x \ln x - \frac{x^2}{2} - 3x

Step 3: Find yzyz

From the equation (2x+3)lnxdx=yz\int (2x + 3) \ln x \, dx = yz, we equate the above expression to yzyz:

yz=x2lnx+3xlnxx223xyz = x^2 \ln x + 3x \ln x - \frac{x^2}{2} - 3x

Step 4: Compare with Options

The expression most closely resembles option (b):

(2x+3)lnx(2x + 3) \ln x

Final Answer:

The correct answer is (b).


Do you want further details or have any questions? Here are 5 related questions to deepen your understanding:

  1. How does integration by parts work, and when is it useful?
  2. Why do we split the integral when integrating functions like (2x+3)lnx(2x + 3) \ln x?
  3. What are other methods to approach integrals involving logarithmic functions?
  4. Could substitution methods have been used here instead of integration by parts?
  5. How can you verify your integral solutions by differentiation?

Tip: When solving integrals involving products of functions, always check if integration by parts or substitution is applicable to simplify the expression.

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 functions

Formulas

∫u dv = uv - ∫v du
∫(2x + 3)lnx dx = ∫2xlnx dx + ∫3lnx dx

Theorems

Integration by parts theorem

Suitable Grade Level

College-level calculus