Math Problem Statement

You're a doctor in mathematics solve this problem

Solution

Let's solve the indicated integral, which is the second one in the image:

xlnxdx\int x \ln x \, dx

This integral requires the method of integration by parts. The formula for integration by parts is:

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

Here, we choose:

  • u=lnxu = \ln x, hence du=1xdxdu = \frac{1}{x} dx
  • dv=xdxdv = x \, dx, hence v=x22v = \frac{x^2}{2}

Now, applying the integration by parts formula:

xlnxdx=x22lnxx221xdx\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx

Simplifying the integral:

xlnxdx=x22lnx12xdx\int x \ln x \, dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx

Now, integrate xdx\int x \, dx:

xdx=x22\int x \, dx = \frac{x^2}{2}

Substituting this into the equation:

xlnxdx=x22lnx12x22+C\int x \ln x \, dx = \frac{x^2}{2} \ln x - \frac{1}{2} \cdot \frac{x^2}{2} + C

xlnxdx=x22lnxx24+C\int x \ln x \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C

Thus, the final answer is:

x22lnxx24+C\boxed{\frac{x^2}{2} \ln x - \frac{x^2}{4} + C}

This is the solution to the second integral. Let me know if you would like a detailed explanation for any other integrals or if you have any questions!

Here are 8 related questions you might find interesting:

  1. How can integration by parts be used for trigonometric integrals?
  2. What is the general method for solving definite integrals?
  3. Can you apply integration by parts multiple times for more complex functions?
  4. How do you integrate functions involving exponential and logarithmic terms together?
  5. What are the applications of integration by parts in physics?
  6. How does the method of substitution differ from integration by parts?
  7. What is the interpretation of the integral xnlnxdx\int x^n \ln x \, dx for different values of nn?
  8. How can you check the correctness of your integration by differentiation?

Tip: When using integration by parts, always choose uu and dvdv such that the resulting integral vduv \, du is simpler to solve than the original.

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Math Problem Analysis

Mathematical Concepts

Integration
Integration by Parts
Logarithmic Functions

Formulas

Integration by Parts formula

Theorems

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Suitable Grade Level

Advanced Mathematics