Evaluating Integrals Of X Arctan(x) And X(ln X)^2 A Comprehensive Guide

Hey guys! Today, we're diving deep into the fascinating world of integral calculus, tackling two intriguing integrals that involve inverse tangent and logarithmic functions. These types of integrals often require a bit of strategic thinking and clever application of integration techniques. So, buckle up and let's get started!

Navigating the Realm of Integration

Before we jump into the specifics, let's briefly touch upon the core concept of integration. At its heart, integration is the reverse process of differentiation. Think of it as finding the area under a curve, accumulating the infinitesimal changes to get a grand total. The integrals we'll be exploring today aren't your run-of-the-mill straightforward integrals; they demand a more nuanced approach, primarily using integration by parts.

Integration by Parts: Our Trusty Tool

Integration by parts is a powerful technique that helps us tackle integrals of products of functions. It's based on the product rule for differentiation and is essentially the reverse of that rule. The formula for integration by parts is:

$$\int u dv = uv - \int v du$$

Where:

• u is a function we choose to differentiate.
• dv is the remaining part of the integrand, which we choose to integrate.
• du is the derivative of u.
• v is the integral of dv.

The key to successfully applying integration by parts lies in the strategic selection of u and dv. We aim to choose u such that its derivative (du) simplifies the integral and dv such that it is easily integrable. This often involves a bit of trial and error, but with practice, you'll get the hang of it!

(a) Tackling the Integral: ∫x arctan(x) dx

Let's dive into our first challenge: evaluating the integral of x * arctan(x) with respect to x. This integral beautifully showcases the power of integration by parts.

Identifying u and dv

The first crucial step is to identify our u and dv. Remember, we want to choose u such that its derivative simplifies the integral. In this case, arctan(x) seems like a good candidate for u because its derivative is a simple rational function. This leaves x dx as our dv.

• Let u = arctan(x), then du = 1/(1 + x^2) dx.
• Let dv = x dx, then v = ∫x dx = (x^2)/2.

Applying Integration by Parts Formula

Now, we plug these into the integration by parts formula:

$$\int x \arctan(x) dx = \arctan(x) * (x^2/2) - \int (x^2/2) * (1/(1 + x^2)) dx$$

This simplifies to:

$$\int x \arctan(x) dx = (x^2/2) \arctan(x) - (1/2) \int x^2/(1 + x^2) dx$$

Conquering the Remaining Integral

We've made progress, but we still have an integral to solve: ∫x^2/(1 + x^2) dx. This integral requires a little algebraic manipulation. We can rewrite the integrand by adding and subtracting 1 in the numerator:

$$\int x^2/(1 + x^2) dx = \int (x^2 + 1 - 1)/(1 + x^2) dx = \int (1 - 1/(1 + x^2)) dx$$

Now, we can split this into two simpler integrals:

$$\int (1 - 1/(1 + x^2)) dx = \int 1 dx - \int 1/(1 + x^2) dx$$

These integrals are standard ones we know:

$$\int 1 dx = x$$

$$\int 1/(1 + x^2) dx = \arctan(x)$$

Putting It All Together

Now, let's substitute these back into our original equation:

$$\int x \arctan(x) dx = (x^2/2) \arctan(x) - (1/2) [x - \arctan(x)] + C$$

Finally, we can simplify this to get our final answer:

$$\int x \arctan(x) dx = (1/2)(x^2 \arctan(x) - x + \arctan(x)) + C$$

Where C is the constant of integration. We've successfully navigated the integral involving the inverse tangent function!

(b) Taming the Integral: ∫x(ln x)^2 dx

Now, let's move on to our second integral: ∫x(ln x)^2 dx. This integral involves the square of the natural logarithm, adding another layer of complexity. However, with our trusty integration by parts, we can conquer this as well.

Strategic Selection of u and dv (Again!)

The key here, as before, is choosing the right u and dv. In this case, (ln x)^2 is a good candidate for u because its derivative will reduce the power of the logarithm. This leaves x dx as our dv.

• Let u = (ln x)^2, then du = 2(ln x)/x dx.
• Let dv = x dx, then v = ∫x dx = (x^2)/2.

Applying Integration by Parts (Like a Boss)

Plugging these into the integration by parts formula, we get:

$$\int x (\ln x)^2 dx = (\ln x)^2 * (x^2/2) - \int (x^2/2) * (2(\ln x)/x) dx$$

Simplifying, we have:

$$\int x (\ln x)^2 dx = (x^2/2) (\ln x)^2 - \int x \ln x dx$$

The Integral Strikes Back! (But We're Ready)

We've simplified things, but we're not quite there yet. We now have another integral to solve: ∫x ln x dx. Guess what? We're going to use integration by parts again! This is a classic example of how sometimes you need to apply a technique multiple times to reach the solution.

Round Two: u and dv Selection

For this integral, we'll choose ln x as our u because its derivative is simple, and x dx as our dv.

• Let u = ln x, then du = (1/x) dx.
• Let dv = x dx, then v = ∫x dx = (x^2)/2.

Integration by Parts: The Sequel

Applying the integration by parts formula again, we get:

$$\int x \ln x dx = (\ln x) * (x^2/2) - \int (x^2/2) * (1/x) dx$$

Simplifying:

$$\int x \ln x dx = (x^2/2) \ln x - (1/2) \int x dx$$

A Familiar Face: ∫x dx

The integral ∫x dx is one we know well:

$$\int x dx = (x^2)/2$$

Putting the Pieces Together (Finally!)

Now, we substitute this back into our equation for ∫x ln x dx:

$$\int x \ln x dx = (x^2/2) \ln x - (1/2) * (x^2/2) = (x^2/2) \ln x - (x^2/4)$$

And now, we substitute this back into our original equation for ∫x (ln x)^2 dx:

$$\int x (\ln x)^2 dx = (x^2/2) (\ln x)^2 - [(x^2/2) \ln x - (x^2/4)] + C$$

The Grand Finale: Simplifying the Result

Finally, we simplify to get our final answer:

$$\int x (\ln x)^2 dx = (x^2/2) (\ln x)^2 - (x^2/2) \ln x + (x^2/4) + C$$

Where C is the constant of integration. Phew! We've successfully tamed the integral involving the square of the natural logarithm!

Key Takeaways and the Journey of Integration

These two integrals beautifully illustrate the power and versatility of integration by parts. The key takeaways are:

• Strategic Selection is Key: Carefully choose your u and dv to simplify the integral.
• Don't Be Afraid to Iterate: Sometimes, you need to apply integration by parts multiple times.
• Algebraic Manipulation is Your Friend: Don't hesitate to use algebraic tricks to rewrite integrands.
• Practice Makes Perfect: The more you practice, the better you'll become at recognizing patterns and applying the appropriate techniques.

Integration, like any mathematical skill, is a journey. There will be challenges, moments of frustration, and, ultimately, the satisfaction of solving a complex problem. So, keep practicing, keep exploring, and keep integrating!

Conclusion

We've successfully navigated two challenging integrals today, showcasing the power of integration by parts and strategic problem-solving. Remember, guys, integration is a journey, and with practice, you'll become masters of this essential calculus technique. Keep exploring, keep integrating, and until next time, happy calculating!