Solving The Indefinite Integral Of (2x+3)(x^2+3x-9)^15 Dx

Hey there, math enthusiasts! Today, we're diving into the fascinating world of indefinite integrals. We've got a fun one to tackle: ∫(2x + 3)(x^2 + 3x - 9)^15 dx. Don't worry, it might look intimidating at first, but we'll break it down step by step and you'll see it's totally manageable. We are going to explore the solution to this integral, making sure everyone can follow along. This is a classic example where u-substitution, a powerful technique in calculus, comes to our rescue. So, grab your pencils and let's get started!

Understanding Indefinite Integrals

Before we jump into the problem, let's quickly recap what indefinite integrals are all about. Indefinite integration, in simple terms, is the reverse process of differentiation. If you have a function, say f(x), its indefinite integral is a function F(x) whose derivative is f(x). The twist? There are infinitely many such functions, all differing by a constant. That's why we always add the '+ c' (the constant of integration) at the end of our indefinite integrals. Think of it like this: if you differentiate x^2 + 5, x^2 - 3, or x^2 + any constant, you'll always get 2x. So, when we integrate 2x, we need to account for all those possible constants, hence the '+ c'.

The indefinite integral is a cornerstone concept in calculus, playing a vital role in solving problems related to areas, volumes, and rates of change. It’s a tool that allows us to undo the process of differentiation, effectively finding the original function from its derivative. For example, imagine you know the velocity of an object at any given time and you want to find its position. Integration is what you need. But it's not just about reversing derivatives; indefinite integrals open doors to solving differential equations, which model a wide array of phenomena in physics, engineering, economics, and more. From the motion of planets to the spread of diseases, indefinite integrals help us understand and predict complex systems. The '+ c' might seem like a small detail, but it’s a crucial reminder that we're not just finding one function, but a family of functions. This constant represents the initial conditions or a baseline value that we might need to determine from additional information. So, when we tackle an indefinite integral, we're not just applying a formula; we're uncovering a fundamental relationship between functions and their rates of change, a relationship that underpins much of modern science and mathematics.

The U-Substitution Technique

Now, let's talk about a nifty trick called u-substitution. This technique is super handy when you have an integral where part of the integrand (the thing you're integrating) is the derivative of another part. In our case, we have (2x + 3) and (x^2 + 3x - 9). Notice anything? The derivative of x^2 + 3x - 9 is 2x + 3, which is exactly what we have in our integral! This is where u-substitution shines. The basic idea behind u-substitution is to simplify the integral by replacing a part of the integrand with a new variable, 'u'. This often transforms a complex integral into a more manageable one. Think of it as a change of variables, like switching from Celsius to Fahrenheit, but for integrals. By choosing the right 'u', we can often unravel the integral and make it easier to solve. It’s a bit like finding the right key to unlock a mathematical puzzle.

To effectively use u-substitution, identifying the correct 'u' is crucial. You're typically looking for a function within the integral whose derivative is also present (up to a constant factor). Once you've chosen your 'u', you find its derivative, 'du', and then try to rewrite the entire integral in terms of 'u' and 'du'. This might involve some algebraic manipulation, but the goal is to eliminate the original variable (in our case, 'x') and replace it with 'u'. After you've integrated with respect to 'u', don't forget the final step: substitute back the original expression in terms of 'x'. This is important because the original problem was in terms of 'x', so your final answer should be too. U-substitution is not just a trick; it’s a fundamental technique that builds upon the chain rule in differentiation. It's a powerful tool in your calculus arsenal, and mastering it will open up a whole new world of integrals you can solve. So, let's see how we can apply it to our problem at hand!

Applying U-Substitution to Our Integral

Okay, let's get our hands dirty with the actual problem. We have the integral ∫(2x + 3)(x^2 + 3x - 9)^15 dx. Remember our u-substitution strategy? We want to find a part of the integrand whose derivative is also present. As we discussed, the derivative of (x^2 + 3x - 9) is (2x + 3), which is perfect! So, let's make the substitution:

Let u = x^2 + 3x - 9

Now, we need to find du, which is the derivative of u with respect to x:

du/dx = 2x + 3

Rearranging this, we get:

du = (2x + 3) dx

Look at that! We have (2x + 3) dx in our original integral, and we've just found that it's equal to du. This is exactly what we wanted. Now we can rewrite our integral entirely in terms of u:

∫(2x + 3)(x^2 + 3x - 9)^15 dx = ∫ u^15 du

See how much simpler that looks? We've transformed a seemingly complex integral into a basic power rule integral. This is the magic of u-substitution at work. By carefully choosing our 'u', we've managed to untangle the original integral and put it in a form that's much easier to handle. The next step is to actually perform the integration with respect to 'u', which is a straightforward application of the power rule. Then, we'll substitute back to get our final answer in terms of 'x'. So, let's move on to the integration step and see how this simplifies even further!

Integrating with Respect to U

Alright, now we have the simplified integral: ∫ u^15 du. This is a classic power rule integral. The power rule states that ∫x^n dx = (x^(n+1))/(n+1) + c, where n is any constant except -1. So, applying the power rule to our integral is a breeze:

∫ u^15 du = (u^(15+1))/(15+1) + c

Simplifying this, we get:

∫ u^15 du = (u^16)/16 + c

That was easy, wasn't it? We've successfully integrated with respect to u. But remember, our original problem was in terms of x, so we're not quite done yet. We need to substitute back to get our answer in terms of x. This is a crucial step because we want our final answer to match the original problem's variable. Think of it as translating back from our 'u' language to the 'x' language. We've done the heavy lifting of integration, but we need to make sure our result is expressed in the context of the original question. So, let's take that 'u' and put back what it represents in terms of 'x'. This will give us the final answer to our indefinite integral.

Substituting Back to X

Okay, time to bring back the x! We know that u = x^2 + 3x - 9. So, we're going to substitute this back into our result:

(u^16)/16 + c = ((x^2 + 3x - 9)^16)/16 + c

And there you have it! We've found the indefinite integral. The final answer is:

∫(2x + 3)(x^2 + 3x - 9)^15 dx = (1/16)(x^2 + 3x - 9)^16 + c

This is the solution we were after. We started with a somewhat intimidating integral, used u-substitution to simplify it, integrated with respect to u, and then substituted back to get our final answer in terms of x. It's like a mathematical journey, and we've reached our destination. But let's not stop here. It's always a good idea to double-check our work, especially in calculus. So, in the next section, we'll verify our result by differentiating it and making sure we get back to our original integrand. This is a great way to build confidence in our solution and reinforce our understanding of the relationship between integration and differentiation.

Verifying the Solution

To verify our solution, we'll differentiate the result we obtained and see if we get back the original integrand. Our solution is:

F(x) = (1/16)(x^2 + 3x - 9)^16 + c

Now, let's differentiate F(x) with respect to x. Remember the chain rule? It's our best friend when differentiating composite functions like this. The chain rule states that if we have a function f(g(x)), its derivative is f'(g(x)) * g'(x). So, we'll apply the chain rule here:

F'(x) = d/dx [(1/16)(x^2 + 3x - 9)^16 + c]

The constant (1/16) comes along for the ride, and the derivative of the constant c is zero. So, we have:

F'(x) = (1/16) * 16 * (x^2 + 3x - 9)^(16-1) * d/dx [x^2 + 3x - 9]

Notice how we've applied the power rule and the chain rule in one step. The exponent 16 comes down, we reduce it by 1, and then we multiply by the derivative of the inside function (x^2 + 3x - 9). Now, let's find that derivative:

d/dx [x^2 + 3x - 9] = 2x + 3

Plugging this back into our expression for F'(x), we get:

F'(x) = (1/16) * 16 * (x^2 + 3x - 9)^15 * (2x + 3)

Simplifying, the 16s cancel out:

F'(x) = (2x + 3)(x^2 + 3x - 9)^15

And guess what? This is exactly our original integrand! This confirms that our solution is correct. We differentiated our result and got back to where we started, which is the hallmark of a correct indefinite integral. Verifying our solution not only gives us confidence in our answer but also reinforces our understanding of the fundamental relationship between differentiation and integration. It's a great practice to adopt whenever you're working with integrals, as it can help you catch any errors and solidify your knowledge.

Conclusion

Awesome job, guys! We've successfully found the indefinite integral of ∫(2x + 3)(x^2 + 3x - 9)^15 dx. We used the powerful technique of u-substitution to simplify the integral, applied the power rule, and then verified our solution by differentiation. The final answer is:

(1/16)(x^2 + 3x - 9)^16 + c

Remember, indefinite integrals might seem tricky at first, but with practice and the right techniques, you can conquer them. U-substitution is a key tool in your calculus arsenal, so make sure you're comfortable with it. Keep practicing, and you'll become an integration pro in no time! And most importantly, have fun with math! It's a beautiful and fascinating subject, and there's always something new to learn and explore.