Math Problem Statement

Solve the integral \( \int \frac{\ln x}{x^2} \, dx \)

Solution

The integral given in the image is:

lnxx2dx\int \frac{\ln x}{x^2} \, dx

To solve this integral, we can use the method of integration by parts. Recall the formula for integration by parts:

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

Let u=lnxu = \ln x, so that du=1xdxdu = \frac{1}{x} \, dx.
Let dv=1x2dxdv = \frac{1}{x^2} \, dx, so that v=1xv = -\frac{1}{x}.

Now apply the formula:

lnxx2dx=uvvdu\int \frac{\ln x}{x^2} \, dx = u v - \int v \, du

Substitute u=lnxu = \ln x, v=1xv = -\frac{1}{x}, and du=1xdxdu = \frac{1}{x} \, dx:

lnxx2dx=(lnx1x)(1x1x)dx\int \frac{\ln x}{x^2} \, dx = \left( \ln x \cdot -\frac{1}{x} \right) - \int \left( -\frac{1}{x} \cdot \frac{1}{x} \right) dx

Simplify:

lnxx2dx=lnxx+1x2dx\int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx

The second integral is straightforward:

1x2dx=x2dx=x1=1x\int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = -x^{-1} = -\frac{1}{x}

Now substitute back:

lnxx2dx=lnxx1x+C\int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} - \frac{1}{x} + C

Final answer:

lnxx2dx=lnxx1x+C\int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} - \frac{1}{x} + C

Let me know if you'd like further clarification or details on this solution.


Here are 5 related questions for further exploration:

  1. How does integration by parts work for different types of functions?
  2. What are some practical applications of logarithmic integrals in physics or engineering?
  3. How can substitution simplify other integrals with logarithmic terms?
  4. How does the constant of integration CC influence the solution?
  5. What happens if we change the power of xx in the denominator?

Tip: When choosing uu in integration by parts, pick the function that simplifies upon differentiation.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration by Parts
Logarithmic Functions

Formulas

Integration by parts formula: \( \int u \, dv = uv - \int v \, du \)

Theorems

Integration by parts

Suitable Grade Level

College Level