Math Problem Statement

Use integration by parts to find ∫(ln x)^2 dx.

Solution

The problem asks to use integration by parts to solve the integral:

(lnx)2dx\int (\ln x)^2 \, dx

We can solve this using the method of integration by parts, which is given by:

udv=uvvdu\int u \, dv = uv - \int v \, du

Step-by-step Solution:

  1. Choose uu and dvdv:

    We set: u=(lnx)2so thatdu=2lnx1xdxu = (\ln x)^2 \quad \text{so that} \quad du = 2 \ln x \cdot \frac{1}{x} \, dx And, dv=dxso thatv=xdv = dx \quad \text{so that} \quad v = x

  2. Apply the Integration by Parts Formula:

    Using the formula udv=uvvdu\int u \, dv = uv - \int v \, du, we have: (lnx)2dx=x(lnx)2x2lnx1xdx\int (\ln x)^2 \, dx = x (\ln x)^2 - \int x \cdot 2 \ln x \cdot \frac{1}{x} \, dx Simplify the second term: x2lnx1xdx=2lnxdx\int x \cdot 2 \ln x \cdot \frac{1}{x} \, dx = \int 2 \ln x \, dx

  3. Solve 2lnxdx\int 2 \ln x \, dx:

    Again, we apply integration by parts for 2lnxdx\int 2 \ln x \, dx.

    Let: u=lnxso thatdu=1xdxu = \ln x \quad \text{so that} \quad du = \frac{1}{x} \, dx And, dv=dxso thatv=xdv = dx \quad \text{so that} \quad v = x

    Now apply the formula: 2lnxdx=2(xlnxx1xdx)\int 2 \ln x \, dx = 2 (x \ln x - \int x \cdot \frac{1}{x} \, dx) Simplify: 2lnxdx=2(xlnxx)=2xlnx2x\int 2 \ln x \, dx = 2 (x \ln x - x) = 2x \ln x - 2x

  4. Substitute Back:

    Now substitute back into the original expression: (lnx)2dx=x(lnx)2(2xlnx2x)+C\int (\ln x)^2 \, dx = x (\ln x)^2 - (2x \ln x - 2x) + C Simplify: (lnx)2dx=x(lnx)22xlnx+2x+C\int (\ln x)^2 \, dx = x (\ln x)^2 - 2x \ln x + 2x + C

Thus, the integral is:

(lnx)2dx=x((lnx)22lnx+2)+C\int (\ln x)^2 \, dx = x \left( (\ln x)^2 - 2 \ln x + 2 \right) + C

Would you like further clarification or additional details on any steps?


Here are some related questions to explore:

  1. How does integration by parts generally work?
  2. Why was u=(lnx)2u = (\ln x)^2 chosen in this case?
  3. What other methods can be used to integrate functions involving lnx\ln x?
  4. How can integration by parts be extended to higher powers of lnx\ln x?
  5. What are the applications of such logarithmic integrals in physics or engineering?

Tip: When applying integration by parts, always choose uu and dvdv carefully. Selecting uu as the part that simplifies upon differentiation often helps streamline the solution.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration by Parts
Logarithmic Functions

Formulas

Integration by Parts Formula: ∫u dv = uv - ∫v du
Logarithmic Differentiation: d(ln x)/dx = 1/x

Theorems

Integration by Parts

Suitable Grade Level

Undergraduate Calculus (University Level)