How do you find the integration of log x?

Finding the integral of the natural logarithm function, specifically ∫ log(x) dx, involves using integration by parts, a technique that is particularly useful for integrating products of functions. Let's break this down step by step.

Understanding Integration by Parts

Integration by parts is based on the product rule for differentiation. The formula is given by:

∫ u dv = uv - ∫ v du

In this formula, you choose parts of your integral to represent u and dv. The goal is to simplify the integral into a more manageable form.

Choosing u and dv

For the integral of log(x), we can make the following choices:

• u = log(x) (which we will differentiate)
• dv = dx (which we will integrate)

Calculating du and v

Next, we need to find du and v:

• To find du, we differentiate u: du = (1/x) dx
• To find v, we integrate dv: v = x

Applying the Integration by Parts Formula

Now we can substitute these values into the integration by parts formula:

∫ log(x) dx = x log(x) - ∫ x (1/x) dx

This simplifies to:

∫ log(x) dx = x log(x) - ∫ 1 dx

Solving the Remaining Integral

The integral of 1 with respect to x is simply x:

∫ 1 dx = x

Now, substituting this back into our equation gives:

∫ log(x) dx = x log(x) - x + C

Here, C represents the constant of integration, which is included because we are dealing with an indefinite integral.

Final Result

Thus, the integral of log(x) is:

∫ log(x) dx = x log(x) - x + C

Example for Clarity

Let’s consider an example to illustrate this result. If you wanted to evaluate the integral from 1 to e (where e is the base of natural logarithms), you would set up:

∫ from 1 to e log(x) dx = [x log(x) - x] from 1 to e

Calculating this, you would find:

• At x = e: e log(e) - e = e(1) - e = 0
• At x = 1: 1 log(1) - 1 = 1(0) - 1 = -1

So, the definite integral evaluates to:

0 - (-1) = 1

This process not only gives you the integral of log(x) but also illustrates how integration by parts can be a powerful tool in calculus. If you have any further questions or need clarification on any step, feel free to ask!