Evaluating The Indefinite Integral Of 6 Divided By X Ln(2x) Using U-Substitution

Hey everyone! Today, we're diving into a fun calculus problem: evaluating the indefinite integral of 6 divided by x times the natural logarithm of 2x. This might sound a bit intimidating at first, but don't worry, we'll break it down step by step using a clever technique called u-substitution. So, grab your pencils, and let's get started!

Understanding the Integral

Before we jump into the solution, let's make sure we understand what we're dealing with. The indefinite integral we need to solve is:

$$\int \frac{6 dx}{x \ln (2 x)}$$

What this means is we're looking for a function whose derivative is 6 divided by (x multiplied by the natural logarithm of 2x). Integrals like this, where we have a function inside another function (in this case, ln(2x) inside the denominator), are often great candidates for u-substitution. This method allows us to simplify the integral by replacing a complex expression with a single variable, making it easier to handle.

The Magic of u-Substitution

The u-substitution technique hinges on the chain rule of differentiation in reverse. The main idea is to choose a part of the integrand (the function we're integrating) as our u, and its derivative should also be present in the integral (possibly up to a constant multiple). This lets us rewrite the integral in terms of u and du, hopefully simplifying things significantly. In our case, the natural choice for u is the natural logarithm of 2x, ln(2x). Let's see why.

Let's set:

$$u = \ln(2x)$$

Now, we need to find the derivative of u with respect to x, which we denote as du/dx. Using the chain rule, the derivative of ln(2x) is:

$$\frac{du}{dx} = \frac{1}{2x} \cdot 2 = \frac{1}{x}$$

Notice something cool? We have 1/x in our original integral! This is exactly what we wanted. Now, we can rewrite this as:

$$du = \frac{1}{x} dx$$

This means that dx/x in our original integral can be directly replaced with du. This is the core of u-substitution. We've managed to connect our chosen u with a part of the original integral.

Rewriting the Integral in Terms of u

Now comes the fun part: rewriting the entire integral in terms of u. Remember our original integral:

$$\int \frac{6 dx}{x \ln (2 x)}$$

We said that u = ln(2x) and du = dx/x. We can directly substitute these into the integral. The constant 6 can be pulled out of the integral, making our lives a bit easier. So, we have:

$$6 \int \frac{dx}{x \ln (2 x)} = 6 \int \frac{1}{\ln(2x)} \cdot \frac{dx}{x}$$

Now, let's substitute u and du:

$$6 \int \frac{1}{u} du$$

Look at that! Our integral has transformed into something much simpler. We've successfully used u-substitution to get rid of the complex ln(2x) term and are left with a basic integral that we know how to solve.

Integrating with Respect to u

The integral of 1/u with respect to u is a standard integral, and it's simply the natural logarithm of the absolute value of u. So, we have:

$$6 \int \frac{1}{u} du = 6 \ln |u| + C$$

Don't forget the + C! This is the constant of integration, which is crucial for indefinite integrals. It represents the fact that the derivative of a constant is zero, so there are infinitely many functions that could have the same derivative. We add C to account for all these possibilities.

Substituting Back for x

We're almost there, but we're not quite done yet. Our answer is currently in terms of u, but we want the answer in terms of our original variable, x. Remember that we defined u as ln(2x). So, we need to substitute this back into our expression:

$$6 \ln |u| + C = 6 \ln |\ln(2x)| + C$$

And that's it! We've successfully evaluated the indefinite integral. The final answer is 6 times the natural logarithm of the absolute value of the natural logarithm of 2x, plus the constant of integration.

The Final Answer

Therefore, the indefinite integral of 6 divided by x ln(2x) is:

$$\int \frac{6 dx}{x \ln (2 x)} = 6 \ln |\ln(2x)| + C$$

This result tells us that any function of the form 6*ln(|ln(2x)|) + C, where C is any constant, will have a derivative of 6/(x ln(2x)).

Key Takeaways

Let's recap what we've learned in this example. U-substitution is a powerful technique for simplifying integrals, especially those involving composite functions. The key steps are:

1. Choose a suitable u: Look for a function within a function, and see if its derivative (or a multiple of it) is also present in the integral.
2. Find du/dx and rewrite as du: Calculate the derivative of u with respect to x, and rearrange the equation to express du in terms of dx.
3. Substitute u and du: Replace the chosen function and its derivative in the original integral with u and du, respectively.
4. Evaluate the simplified integral: Integrate the new integral with respect to u.
5. Substitute back for x: Replace u with its original expression in terms of x.
6. Add the constant of integration: Don't forget the + C for indefinite integrals!

By following these steps, you can tackle a wide range of integrals that might initially seem challenging. U-substitution is a fundamental tool in calculus, and mastering it will significantly improve your integration skills.

Practice Makes Perfect

The best way to get comfortable with u-substitution is to practice. Try working through similar examples, and don't be afraid to experiment with different choices for u. Sometimes, the first choice might not work, but that's okay! It's part of the learning process. Keep trying, and you'll get the hang of it. For example, you could try integrals like:

• $\int \frac{2x}{x^2 + 1} dx$
• $\int x \cos(x^2) dx$
• $\int \frac{e^x}{e^x + 1} dx$

These integrals all lend themselves well to u-substitution. Give them a shot, and see how you do!

Conclusion

So, there you have it! We've successfully evaluated the indefinite integral of 6 divided by x ln(2x) using the u-substitution method. We saw how choosing the right u can dramatically simplify an integral, and we learned the key steps involved in this powerful technique. Remember, practice is key, so keep working at it, and you'll become an integration master in no time! Keep exploring the fascinating world of calculus, and I'll see you in the next problem!