Antiderivative of x ln x: Understanding Integration by Parts

Antiderivative of x ln x: Understanding Integration by Parts

In this article, we will explore the antiderivative of the function x ln x. We'll start with a detailed explanation of the integration process using integration by parts and provide a step-by-step solution, complete with relevant mathematical justifications. By the end of this guide, you will have a thorough understanding of how to handle such integrals.

Introduction to Integration by Parts

Integration by parts is a technique used to integrate the product of two functions. It is derived from the product rule of differentiation and is expressed as:

[int u , dv uv - int v , du]

Here, (u) and (dv) are functions of (x), and (du) and (v) are their respective derivatives and anti-derivatives.

Step-by-Step Solution for the Antiderivative of x ln x

Let's begin with the integral:

[int x ln x , dx]

We will use integration by parts to solve this integral. First, we need to choose appropriate functions for (u) and (dv).

Step 1: Choosing u and dv

Recall the integration by parts formula:

[int u , dv uv - int v , du]

In this case, let:

[u ln x]

[dv x , dx]

Then, we need to find (du) and (v):

[du frac{1}{x} , dx]

[v frac{x^2}{2}] (since (int x , dx frac{x^2}{2}))

Substitute (u), (v), (du), and (dv) into the integration by parts formula:

[int x ln x , dx ln x cdot frac{x^2}{2} - int frac{x^2}{2} cdot frac{1}{x} , dx]

Simplify the expression:

[ frac{x^2}{2} ln x - int frac{x}{2} , dx]

Integrate (frac{x}{2}):

[int frac{x}{2} , dx frac{1}{2} int x , dx frac{x^2}{4}]

So, the integral becomes:

[frac{x^2}{2} ln x - frac{x^2}{4} C] (where (C) is the constant of integration)

Therefore, the antiderivative of (x ln x) is:

[boxed{frac{x^2}{2} ln x - frac{x^2}{4} C}]

Conclusion

Through the use of integration by parts, we have successfully found the antiderivative of the function x ln x. Understanding and applying this technique to other similar integrals can significantly enhance your problem-solving skills in calculus. Practice and familiarity with integration by parts are key to leveraging it effectively in more complex scenarios.

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