Evaluating The Integral Of (e^x)/(9e^(2x) + 25) A Step-by-Step Solution

Hey guys! Today, we're diving into the fascinating world of calculus to tackle a rather interesting integral. We're going to break down the process of evaluating the improper integral $\int_{\ln 5 / 3}^{\infty} \frac{e^x}{9 e^{2 x}+25} d x$. This type of problem often appears in calculus courses and can seem daunting at first, but fear not! By carefully applying the right techniques, we can solve it together. So, grab your thinking caps, and let's get started!

Understanding the Problem: Why is this Integral Special?

Before we jump into the calculations, let's take a moment to understand what makes this integral a bit different. First and foremost, it's an improper integral because it has an infinite limit of integration. This means we're trying to find the area under the curve of the function $\frac{e^x}{9 e^{2 x}+25}$ as $x$ extends infinitely to the right.

Secondly, the integrand itself, $\frac{e^x}{9 e^{2 x}+25}$, is a bit complex. It's not a straightforward function that we can integrate directly using basic rules. This hints that we'll need to employ a clever substitution technique to simplify the expression. Identifying these key characteristics is the first step towards successfully solving the problem. We need to acknowledge the infinite limit and the complex integrand to choose the right approach. Think of it as diagnosing the problem before prescribing the solution – a fundamental principle in mathematics and many other fields!

To further grasp the challenge, consider the behavior of the function as $x$ approaches infinity. The exponential term $e^x$ grows rapidly, and so does $e^{2x}$. This means the denominator, $9e^{2x} + 25$, also becomes very large. The question then becomes: does the function decay rapidly enough to ensure that the area under the curve remains finite? Or does the area grow infinitely large, leading to a divergent integral? These are the questions we'll be answering as we work through the solution. Remember, guys, calculus is all about understanding limits and how functions behave in extreme conditions, and this integral is a perfect example of that.

The Substitution Game: Simplifying the Integral

The heart of solving this integral lies in recognizing a suitable substitution. When we see expressions like $e^x$ and $e^{2x}$ together, a lightbulb should go off in our heads: u-substitution! Let's make the substitution $u = 3e^x$. Why this substitution? Well, notice that the derivative of $u$ with respect to $x$ is $du/dx = 3e^x$, which means $du = 3e^x dx$. This is great because we have an $e^x dx$ term in our integral, just waiting to be replaced!

By choosing $u = 3e^x$, we're aiming to transform the integral into a form that we can readily recognize and integrate. This is a common strategy in calculus – to massage the expression into a familiar shape. Now, let's see how this substitution affects the rest of the integral. First, we need to rewrite the $e^{2x}$ term. Since $u = 3e^x$, we can square both sides and get $u^2 = 9e^{2x}$. This is exactly what we have in the denominator! So, we can replace $9e^{2x}$ with $u^2$. Next, we need to deal with the $e^x dx$ term. From our earlier calculation, we know that $du = 3e^x dx$. Therefore, e^x dx = rac{1}{3} du. This is the piece that makes the substitution work so beautifully – we've managed to replace the entire $e^x dx$ term with a simple expression involving $du$.

But hold on, we're not done yet! We also need to change the limits of integration. Our original integral was with respect to $x$, but now we're integrating with respect to $u$. So, we need to find the corresponding $u$ values for our original $x$ limits. The lower limit was $x = \ln(5/3)$. Plugging this into our substitution, we get $u = 3e^{\ln(5/3)} = 3(5/3) = 5$. For the upper limit, $x = \infty$, we have $u = 3e^{\infty}$. As $x$ approaches infinity, $e^x$ also approaches infinity, so $u$ approaches infinity as well. Therefore, our new limits of integration are $u = 5$ and $u = \infty$. See how carefully we've handled every detail? This meticulous approach is crucial for avoiding errors and ensuring the correctness of our solution.

Putting it all together, our integral now looks like this: $\int_{5}^{\infty} \frac{(1/3)}{u^2+25} du$. Doesn't that look much simpler? We've successfully transformed the integral into a more manageable form. Now, we're ready to tackle the next step: recognizing the standard integral form.

Spotting the Pattern: The Arctangent Savior

The transformed integral, $\int_{5}^{\infty} \frac{(1/3)}{u^2+25} du$, is a classic form that should ring a bell for anyone familiar with integral calculus. It closely resembles the integral of the arctangent function. Remember the formula: $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \arctan(\frac{x}{a}) + C$, where $C$ is the constant of integration.

In our case, we have $u^2 + 25$, which can be written as $u^2 + 5^2$. So, $a = 5$. This is a crucial observation! Recognizing this pattern allows us to directly apply the arctangent formula. It's like having the right key to unlock the solution. We just need to carefully match the pieces.

Before we apply the formula, let's factor out the constant $(1/3)$ from the integral: $\frac{1}{3} \int_{5}^{\infty} \frac{1}{u^2+25} du$. Now, we can clearly see the arctangent form. Applying the formula, we get: $\frac{1}{3} \cdot \frac{1}{5} \arctan(\frac{u}{5})$. Don't forget the factor of $(1/a)$, which in this case is $(1/5)$. It's these small details that often make the difference between a correct and incorrect solution. Pay close attention to the constants!

So, the antiderivative of our integrand is $\frac{1}{15} \arctan(\frac{u}{5})$. Now, we need to evaluate this antiderivative at the limits of integration, $5$ and $\infty$. This is where the concept of improper integrals comes into play. We can't directly plug in infinity, so we need to use a limit.

Taming Infinity: Evaluating the Improper Integral

Since we're dealing with an improper integral, we need to use limits to evaluate it correctly. Remember, infinity isn't a number; it's a concept representing unbounded growth. So, to evaluate the integral $\frac{1}{3} \int_{5}^{\infty} \frac{1}{u^2+25} du$, we'll replace the upper limit of integration with a variable, say $t$, and then take the limit as $t$ approaches infinity. This gives us: $\frac{1}{3} \lim_{t \to \infty} \int_{5}^{t} \frac{1}{u^2+25} du$.

We already found the antiderivative to be $\frac{1}{15} \arctan(\frac{u}{5})$. So, we can rewrite the expression as: $\frac{1}{15} \lim_{t \to \infty} [\arctan(\frac{u}{5})]_{5}^{t}$. Now, we need to evaluate the antiderivative at the limits $t$ and $5$ and subtract: $\frac{1}{15} \lim_{t \to \infty} [\arctan(\frac{t}{5}) - \arctan(\frac{5}{5})]$. This simplifies to: $\frac{1}{15} \lim_{t \to \infty} [\arctan(\frac{t}{5}) - \arctan(1)]$.

Now comes the crucial part: evaluating the limit. As $t$ approaches infinity, $\frac{t}{5}$ also approaches infinity. We need to know what happens to the arctangent function as its argument goes to infinity. Recall that the arctangent function, $\arctan(x)$, represents the angle whose tangent is $x$. As $x$ approaches infinity, the angle approaches $\frac{\pi}{2}$ (or 90 degrees). So, $\lim_{t \to \infty} \arctan(\frac{t}{5}) = \frac{\pi}{2}$.

We also know that $\arctan(1) = \frac{\pi}{4}$ (since the tangent of $\frac{\pi}{4}$ is 1). Plugging these values back into our expression, we get: $\frac{1}{15} [\frac{\pi}{2} - \frac{\pi}{4}]$. Simplifying this, we get: $\frac{1}{15} [\frac{2\pi}{4} - \frac{\pi}{4}] = \frac{1}{15} \cdot \frac{\pi}{4} = \frac{\pi}{60}$.

The Grand Finale: The Integral's Value

After all the careful steps and calculations, we've arrived at the final answer! The integral $\int_{\ln 5 / 3}^{\infty} \frac{e^x}{9 e^{2 x}+25} d x$ converges, and its value is $\frac{\pi}{60}$. This is a fantastic result! We've successfully navigated an improper integral, used a clever substitution, recognized a standard integral form, and carefully evaluated limits.

Let's take a moment to appreciate the journey. We started with a seemingly complex integral, but by breaking it down into smaller, manageable steps, we were able to conquer it. This is a key lesson in mathematics and problem-solving in general. Don't be intimidated by the size of the problem; focus on the individual steps, and you'll get there.

Moreover, this problem showcases the power of calculus. We were able to find the exact area under a curve that extends infinitely! This is a testament to the elegance and usefulness of calculus in modeling and understanding the world around us. So, guys, keep practicing, keep exploring, and keep enjoying the beauty of mathematics! You've got this!