Integrate (1-x)/x: A Simple Calculus Solution

Hey guys! Today, we're diving into a fun little integral problem. We'll be using a super fundamental result and a simple property to solve it. Get ready to dust off your calculus hats, and let's get started!

The Fundamental Result and Our Goal

We're going to lean heavily on this cornerstone of integral calculus:

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

This basically says that the integral of 1 over a variable u (with respect to u) is the natural logarithm of the absolute value of u, plus a constant of integration (we always need that C, remember?). It's crucial we keep this in mind. This is our go-to formula for anything that looks remotely like 1 over something.

Our mission, should we choose to accept it (and we do!), is to evaluate the following integral:

$$\int \frac{1-x}{x} dx$$

At first glance, it might not look like it fits our fundamental result directly. But don't worry! We have a secret weapon: the linearity of the integral. This property is super useful for breaking down complex integrals into simpler, manageable pieces. Let's see how it works.

Leveraging Linearity: Breaking Down the Integral

The linearity of the integral is a fancy way of saying that we can split integrals over sums and differences, and we can pull out constant factors. Mathematically, it means these two things:

1. $$\int [f(x) + g(x)] dx = \int f(x) dx + \int g(x) dx$$

2. \int k \cdot f(x) dx = k \int f(x) dx$, where *k* is a constant.

These rules are absolute game-changers when facing a complicated integral. They allow us to dissect the problem into smaller, more digestible chunks. Now, how does this help us with our integral? Well, let's rewrite the integrand (the function inside the integral) to make things clearer:

$$\frac{1-x}{x} = \frac{1}{x} - \frac{x}{x} = \frac{1}{x} - 1$$

See what we did there? We split the fraction into two simpler terms. This is a classic algebraic manipulation that often paves the way for solving integrals. Now, our integral looks like this:

$$\int \frac{1-x}{x} dx = \int \left( \frac{1}{x} - 1 \right) dx$$

Now we can apply the linearity property! We split the integral over the difference:

$$\int \left( \frac{1}{x} - 1 \right) dx = \int \frac{1}{x} dx - \int 1 dx$$

Boom! Look at that. We've transformed one integral into two, each of which is much easier to handle. The magic of linearity, guys!

Solving the Simplified Integrals

Now, let's tackle each of these integrals separately. The first one, $\int \frac{1}{x} dx$, should look incredibly familiar. It's precisely in the form of our fundamental result!

Applying our rule, we get:

$$\int \frac{1}{x} dx = \ln |x| + C_1$$

Notice the C₁? We're keeping track of the constants of integration for each individual integral. We'll combine them at the very end. Now, let's move on to the second integral:

$$\int 1 dx$$

This one's a classic too. What function, when differentiated, gives us 1? Well, that's just x! So,

$$\int 1 dx = x + C_2$$

Again, we have a constant of integration, C₂. We're almost there! We've successfully integrated both parts. It's like we're cooking, and all the ingredients are prepped and ready to go.

Putting It All Together: The Grand Finale

Now comes the exciting part: combining our results! Remember, we had:

$$\int \frac{1-x}{x} dx = \int \frac{1}{x} dx - \int 1 dx$$

And we found:

$$\int \frac{1}{x} dx = \ln |x| + C_1$$

$$\int 1 dx = x + C_2$$

Substituting these back into our main equation, we get:

$$\int \frac{1-x}{x} dx = (\ln |x| + C_1) - (x + C_2)$$

Let's simplify this a bit:

$$\int \frac{1-x}{x} dx = \ln |x| - x + C_1 - C_2$$

Now, here's a neat trick. We have two constants of integration, C₁ and C₂. Since they're both arbitrary constants, their difference (C₁ - C₂) is also just an arbitrary constant! We can replace it with a single constant, let's call it C:

$$\int \frac{1-x}{x} dx = \ln |x| - x + C$$

And there you have it! We've successfully evaluated the integral. Awesome job, guys! We took a seemingly tricky integral, broke it down using linearity, applied a fundamental result, and combined the pieces to get our final answer.

Wrapping Up: Key Takeaways

Let's recap what we've learned. This problem perfectly illustrates the power of a few key concepts in integral calculus:

• Fundamental Results: Knowing basic integrals like $\int \frac{1}{u} du = \ln |u| + C$ is crucial. They're the building blocks for solving more complex integrals.
• Linearity of the Integral: This property is a lifesaver. It allows us to split integrals over sums and differences, and to pull out constant factors. This often transforms a difficult integral into a set of easier ones.
• Algebraic Manipulation: Sometimes, a little algebra is all you need to simplify an integrand and make it integrable. In our case, splitting the fraction $rac{1-x}{x}$ was the key to unlocking the solution.
• Constants of Integration: Don't forget those C's! They're an essential part of the indefinite integral.

So, next time you're faced with an integral that looks a bit daunting, remember these techniques. Break it down, look for familiar patterns, and don't be afraid to use a little algebraic trickery. You've got this!

Keep practicing, keep exploring, and have fun with calculus! You'll be mastering integrals in no time. And hey, if you have any questions or want to try another integral problem, let me know in the comments below. Let's keep the learning going!