Solving ∫1^e [1/(x(1+ln X)^2)] Dx: A Step-by-Step Guide

Hey math enthusiasts! Ever stumbled upon an integral that looks intimidating but is actually quite manageable? Today, we're diving deep into solving the definite integral ∫1^e [1/(x(1+ln x)^2)] dx. This is a classic calculus problem that combines a bit of algebraic manipulation with a clever substitution. So, buckle up, grab your notebooks, and let's get started!

Breaking Down the Integral

At first glance, the integral ∫1^e [1/(x(1+ln x)^2)] dx might seem a bit daunting. You've got a fraction, a natural logarithm, and a squared term all hanging out together. But don't worry, we're going to break it down step-by-step. The key to solving this integral lies in recognizing a suitable substitution. Often in calculus, spotting the right substitution can turn a complex problem into a straightforward one. In this case, we'll use what's known as a u-substitution. This technique involves replacing a part of the integrand (the function inside the integral) with a new variable, u, to simplify the expression. This not only makes the integral easier to handle but also often reveals a more recognizable form that we can integrate using standard rules. It's like finding a secret code that unlocks the solution! We'll carefully choose u so that its derivative is also present in the integral, allowing us to neatly rewrite the entire expression in terms of u. This is where the magic happens, transforming a potentially messy problem into a smooth, solvable one. Trust me, guys, once you get the hang of u-substitution, you'll start seeing opportunities to use it everywhere in calculus!

Identifying the Right Substitution

Okay, so the million-dollar question is: what should we substitute? Look closely at the integral: ∫1^e [1/(x(1+ln x)^2)] dx. Notice that we have a ln x term inside the denominator. And remember, the derivative of ln x is 1/x, which is also present in the integral! This is a major clue that a u-substitution involving 1 + ln x might be the way to go. So, let's try setting u = 1 + ln x. This is a classic strategy in calculus—when you see a function and its derivative (or a multiple of its derivative) within the integral, it’s often a sign that a well-chosen substitution can simplify things dramatically. The logic here is that by making this substitution, we can transform the complex expression in the denominator into something much simpler, making the integral as a whole more approachable. Now, why 1 + ln x instead of just ln x? Including the 1 is a smart move because it will disappear when we take the derivative, further cleaning up the expression. It’s these little tricks and insights that can really make a difference in tackling these kinds of problems, and with a little practice, you'll start spotting them like a pro!

Performing the Substitution

Now that we've identified our substitution (u = 1 + ln x), let's put it into action! The next step is to find the differential du. Remember, the differential is just the derivative of u with respect to x, multiplied by dx. So, if u = 1 + ln x, then du/dx = 1/x. Multiplying both sides by dx, we get du = (1/x) dx. Look familiar? This is exactly the term we have in our original integral! This is what makes u-substitution so elegant—it allows us to neatly replace parts of the integral with our new variable and its differential. Now, let's rewrite the original integral in terms of u. We have 1/(x(1+ln x)^2) dx, which can be rearranged as (1/(1+ln x)^2) * (1/x) dx. Substituting u = 1 + ln x and du = (1/x) dx, our integral transforms into ∫(1/u^2) du. This looks way simpler, doesn't it? We've effectively turned a complex expression into a basic power rule integral. This is the power of substitution, guys—it's like magic! By choosing the right substitution, we've made the problem significantly more manageable. Now, we’re in a position to apply standard integration techniques and actually solve the integral. High five for that!

Changing the Limits of Integration

But hold on, we're not quite done with the substitution process! Because we're dealing with a definite integral (one with limits of integration), we need to adjust those limits to match our new variable, u. Our original limits are in terms of x: from x = 1 to x = e. We need to find the corresponding u values. Remember, u = 1 + ln x. So, when x = 1, u = 1 + ln 1 = 1 + 0 = 1. And when x = e, u = 1 + ln e = 1 + 1 = 2. So, our new limits of integration are from u = 1 to u = 2. Why is this step so crucial? Well, if we didn't change the limits, we'd have to substitute back for x at the end after integrating with respect to u. That's totally doable, but it adds an extra step and potential for errors. By changing the limits now, we can directly evaluate the integral in terms of u and get our final answer without any back-substitution hassle. It's all about working smarter, not harder, right? Plus, it keeps the flow of the problem nice and clean. So, always remember when you're doing a definite integral with substitution: change those limits! It's a small step that makes a big difference in the long run.

Evaluating the Integral

Alright, we've made a fantastic transformation! Our integral has gone from ∫1^e [1/(x(1+ln x)^2)] dx to ∫1^2 (1/u^2) du. This is a much friendlier integral to deal with. To make it even clearer, let's rewrite 1/u^2 as u^(-2). Now we have ∫1^2 u^(-2) du. This is a classic power rule integral. Remember the power rule for integration? It says that ∫x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration. In our case, n = -2. Applying the power rule, we get ∫ u^(-2) du = (u^(-1))/(-1) + C = -1/u + C. But since we're dealing with a definite integral, we don't need to worry about the constant of integration C because it will cancel out when we evaluate the limits. Now, we need to evaluate -1/u from u = 1 to u = 2. This means we'll plug in the upper limit (u = 2) and subtract the result of plugging in the lower limit (u = 1). So, we have [-1/2] - [-1/1] = -1/2 + 1 = 1/2. And there you have it! The value of the integral is 1/2. Isn’t it satisfying when a seemingly complex problem boils down to such a neat and tidy answer? This whole process really shows the power of strategic substitution and the fundamental rules of calculus. We've conquered this integral, guys!

Final Answer and Conclusion

So, after all that work, we've arrived at our final answer: ∫1^e [1/(x(1+ln x)^2)] dx = 1/2. Woohoo! Give yourselves a pat on the back for making it through this journey. We took a seemingly complicated integral, used a clever u-substitution, changed the limits of integration, applied the power rule, and emerged victorious with a numerical answer. This is what calculus is all about, guys! It's about taking complex problems and breaking them down into manageable steps. By recognizing patterns (like the ln x and 1/x relationship), choosing the right techniques (like u-substitution), and paying attention to detail (like changing the limits), you can tackle almost any integral that comes your way. Remember, practice makes perfect. The more you work through these types of problems, the more comfortable and confident you'll become. So, keep those pencils moving, keep exploring the world of calculus, and keep having fun with math! You've got this!