Solving ∫x⁶√(⁵x⁷+7) Dx A Step-by-Step Guide With U-Substitution

Hey there, math enthusiasts! Today, we're diving deep into the fascinating world of indefinite integrals, specifically tackling the integral ∫x⁶√(⁵x⁷+7) dx. This may look intimidating at first glance, but don't worry, we'll break it down step by step. This guide will equip you with the knowledge and confidence to conquer similar integration problems. Our focus will be on using u-substitution, a powerful technique for simplifying complex integrals. Let's get started!

Understanding U-Substitution: The Key to Simplification

Before we jump into the problem, let's quickly review the concept of u-substitution. U-substitution, also known as variable substitution, is a technique used to simplify integrals by replacing a complex expression within the integral with a single variable, 'u'. The goal is to transform the integral into a simpler form that we can easily integrate using basic integration rules. This method is particularly effective when the integrand contains a composite function, like our √(⁵x⁷+7) term. To effectively employ u-substitution, you need to identify a suitable 'u' and its derivative, 'du', within the integral. The key is to choose a 'u' such that its derivative is also present in the integral, or can be easily obtained by adjusting a constant factor. By substituting 'u' and 'du', we can often transform a seemingly complicated integral into a straightforward one. Remember, the ultimate aim is to make the integral solvable using the fundamental rules of integration. The u-substitution method can be seen as the reverse of the chain rule in differentiation, which is why it's so effective for handling composite functions. It's like finding a hidden key that unlocks the solution to the integral. The technique allows us to simplify the integral, making it much easier to handle and solve. Once we find the integral in terms of u, we need to remember to substitute back in terms of x to get the final answer. This method not only makes integration easier but also provides a deeper understanding of how different parts of the integral relate to each other. So, let's keep this core concept in mind as we solve our problem.

Identifying 'u' and 'du' in ∫x⁶√(⁵x⁷+7) dx

Alright, let's get our hands dirty with our integral: ∫x⁶√(⁵x⁷+7) dx. The first crucial step in u-substitution is identifying the 'u'. Looking at the integral, we need to pinpoint a part of the expression whose derivative is also present (or can be manipulated to be present). In this case, the expression inside the fifth root, x⁷+7, looks promising. Why? Because its derivative will involve a term with x⁶, which we already have in our integral! Let's make our choice:

• u = x⁷ + 7

Now, let's find 'du', which is the derivative of 'u' with respect to 'x', multiplied by 'dx':

• du = (d/dx)(x⁷ + 7) dx
• du = 7x⁶ dx

Notice anything exciting? We have x⁶ dx in our integral! We're on the right track. However, our 'du' has a 7 that our original integral doesn't. No sweat! We can easily adjust for this. We need to isolate x⁶ dx in the 'du' equation. To do this, we divide both sides of the du equation by 7:

• du / 7 = x⁶ dx

Excellent! Now we have both 'u' and 'du' (or rather, du/7) expressed in terms of 'x' and 'dx'. This is the key to transforming our integral. Identifying the correct u and subsequently finding du is often the most challenging aspect of u-substitution, but with practice, it becomes second nature. The goal is to simplify the integral, and choosing the right u is the first major step in achieving that simplification. By carefully observing the integrand, we can make an educated guess for u, and then verify our choice by calculating du. This process often involves some trial and error, but the ability to recognize suitable substitutions is a valuable skill in calculus. So, let's move forward and see how this substitution helps us solve the integral.

Performing the Substitution: Transforming the Integral

Okay, guys, we've identified our 'u' (x⁷ + 7) and 'du' (7x⁶ dx), and we've even massaged 'du' into a form we can directly substitute (du/7 = x⁶ dx). Now comes the fun part: actually substituting these into our original integral, ∫x⁶√(⁵x⁷+7) dx. Let's replace x⁷ + 7 with 'u' and x⁶ dx with 'du/7'. Our integral transforms into:

∫√(⁵u) (du/7)

See how much simpler that looks? The complex expression inside the fifth root is now just a single variable, 'u'. We can even pull the constant 1/7 out of the integral, making it even cleaner:

(1/7) ∫√(⁵u) du

Now, to make things even clearer for integration, let's rewrite the fifth root as a fractional exponent:

(1/7) ∫u^(1/5) du

This is a basic power rule integral! Remember the power rule for integration? ∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C, where C is the constant of integration. We've successfully transformed our original complex integral into a simple, manageable form using u-substitution. This step-by-step transformation is a testament to the power of u-substitution. By carefully choosing our u and du, we've been able to simplify the integral, making it solvable using basic integration rules. The substitution process not only simplifies the integral but also provides a clearer understanding of its structure. It allows us to see the underlying simplicity that was hidden within the original complex expression. So, let's proceed with the integration and see how this transformation leads us to the solution. We have successfully paved the way for the final steps of solving the integral.

Integrating and Back-Substituting: Finding the Solution

Alright, we've got our simplified integral: (1/7) ∫u^(1/5) du. Now it's time to integrate! Using the power rule for integration (∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C), we get:

(1/7) * [u^(1/5 + 1) / (1/5 + 1)] + C

Let's simplify the exponents and fractions:

(1/7) * [u^(6/5) / (6/5)] + C

To divide by a fraction, we multiply by its reciprocal:

(1/7) * (5/6) * u^(6/5) + C

Multiply the fractions:

(5/42) * u^(6/5) + C

We're almost there! We've integrated with respect to 'u', but remember, our original problem was in terms of 'x'. We need to back-substitute our 'u' (x⁷ + 7) back into the equation:

(5/42) * (x⁷ + 7)^(6/5) + C

And there you have it! We've successfully solved the indefinite integral. This final step of back-substitution is crucial. It ensures that our answer is expressed in terms of the original variable, 'x'. The constant of integration, 'C', is a reminder that there are infinitely many antiderivatives for a given function. Each value of 'C' corresponds to a different vertical shift of the antiderivative. This completes the process of u-substitution, showcasing its effectiveness in solving complex integrals. By breaking down the problem into smaller, manageable steps, we were able to successfully navigate the integration process. So, we've not only found the solution but also gained a deeper understanding of the techniques involved. Now, let's recap the entire process to solidify our understanding.

Recap: Mastering U-Substitution

Let's quickly recap the steps we took to solve the indefinite integral ∫x⁶√(⁵x⁷+7) dx using u-substitution:

1. Identify 'u': We chose u = x⁷ + 7 because its derivative would involve x⁶, which was present in the integral.
2. Find 'du': We calculated du = 7x⁶ dx.
3. Adjust 'du': We solved for x⁶ dx to get du/7 = x⁶ dx.
4. Substitute: We replaced x⁷ + 7 with 'u' and x⁶ dx with 'du/7' in the original integral.
5. Simplify: We rewrote the integral as (1/7) ∫u^(1/5) du.
6. Integrate: We used the power rule to integrate, resulting in (1/7) * (5/6) * u^(6/5) + C.
7. Back-substitute: We replaced 'u' with x⁷ + 7 to get the final answer: (5/42) * (x⁷ + 7)^(6/5) + C.

By following these steps, you can tackle a wide range of integrals using u-substitution. Remember, the key is to practice and develop your intuition for choosing the right 'u'. The more problems you solve, the better you'll become at recognizing patterns and applying this powerful technique. U-substitution is a fundamental tool in integral calculus, and mastering it will greatly enhance your problem-solving abilities. This method allows us to simplify complex integrals by introducing a new variable that makes the integration process more manageable. It's like having a secret weapon in your calculus arsenal! So, keep practicing, keep exploring, and keep mastering the art of u-substitution.

Practice Makes Perfect: Further Exploration

To truly master u-substitution, it's crucial to practice with a variety of problems. Try working through similar integrals with different functions and exponents. Experiment with different choices for 'u' to see how they affect the complexity of the integral. Some problems may have multiple possible substitutions, while others may require a bit more algebraic manipulation before u-substitution can be applied. Don't be afraid to make mistakes – they're a valuable part of the learning process. Each problem you solve will help you refine your skills and develop a deeper understanding of u-substitution. Consider exploring integrals involving trigonometric functions, exponential functions, and logarithmic functions. These types of integrals often lend themselves well to u-substitution, and working through them will broaden your experience with the technique. Remember, the goal is not just to find the answer, but to understand the underlying principles and develop a systematic approach to solving integration problems. The more you practice, the more confident and proficient you'll become in applying u-substitution and other integration techniques. So, grab your textbook, find some practice problems, and start exploring the fascinating world of integral calculus!

I hope this comprehensive guide has helped you understand and master the technique of u-substitution for solving indefinite integrals. Keep practicing, and you'll be conquering complex integrals in no time!