Solving The Indefinite Integral Of (36√[5]{x^4} - 18x^2 - 8 + 17/√{x}) Dx

Hey guys! Today, we're diving into the exciting world of indefinite integrals. Specifically, we're going to tackle the integral of a somewhat complex function. Don't worry, we'll break it down step-by-step so it's super easy to follow. Our mission, should we choose to accept it (and we totally do!), is to find the indefinite integral of this beast:

$$\int\left(36 \sqrt[5]{x^4}-18 x^2-8+\frac{17}{\sqrt{x}}\right) d x$$

So, grab your thinking caps and let's get started!

Breaking Down the Integral: A Step-by-Step Guide

To solve this indefinite integral, we'll use the power of the power rule and the linearity of integrals. These are our trusty tools for this adventure. The linearity of integrals allows us to split the integral of a sum (or difference) into the sum (or difference) of individual integrals. This is incredibly helpful because it lets us deal with each term separately. The power rule, on the other hand, helps us integrate terms of the form x^n. Remember, the power rule states that:

$$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$

where n is any real number except -1, and C is the constant of integration. That constant is super important because it reminds us that there are infinitely many possible antiderivatives, each differing by a constant. Let's dive into applying these concepts.

Step 1: Splitting the Integral

First up, we use the linearity of integrals to split our integral into manageable chunks:

$$\int\left(36 \sqrt[5]{x^4}-18 x^2-8+\frac{17}{\sqrt{x}}\right) d x = 36\int x^{\frac{4}{5}} dx - 18\int x^2 dx - 8\int dx + 17\int x^{-\frac{1}{2}} dx$$

See? Much less intimidating already! We've essentially broken the original integral into four smaller, easier-to-handle integrals. This is a crucial step in simplifying complex integrals. We've also pulled the constants (36, -18, -8, and 17) outside their respective integral signs. This is another application of the linearity property, and it makes our calculations cleaner.

Step 2: Applying the Power Rule

Now, let's tackle each integral individually using the power rule. Remember, we add 1 to the exponent and then divide by the new exponent. Let's start with the first term:

Integrating 36∫x^(4/5) dx

Applying the power rule, we get:

$$36\int x^{\frac{4}{5}} dx = 36 \cdot \frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} + C_1 = 36 \cdot \frac{x^{\frac{9}{5}}}{\frac{9}{5}} + C_1 = 36 \cdot \frac{5}{9} x^{\frac{9}{5}} + C_1 = 20x^{\frac{9}{5}} + C_1$$

Integrating -18∫x^2 dx

Next, we integrate the second term:

$$-18\int x^2 dx = -18 \cdot \frac{x^{2+1}}{2+1} + C_2 = -18 \cdot \frac{x^3}{3} + C_2 = -6x^3 + C_2$$

Integrating -8∫dx

Now, let's integrate the third term. Remember that ∫dx is the same as ∫1 dx, which is the same as ∫x^0 dx. Applying the power rule:

$$-8\int dx = -8\int x^0 dx = -8 \cdot \frac{x^{0+1}}{0+1} + C_3 = -8x + C_3$$

Integrating 17∫x^(-1/2) dx

Finally, let's integrate the fourth term:

$$17\int x^{-\frac{1}{2}} dx = 17 \cdot \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + C_4 = 17 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C_4 = 17 \cdot 2x^{\frac{1}{2}} + C_4 = 34\sqrt{x} + C_4$$

Step 3: Putting It All Together

Now that we've integrated each term individually, it's time to combine them. We'll add up all the results and combine the constants of integration into a single constant, C:

$$\int\left(36 \sqrt[5]{x^4}-18 x^2-8+\frac{17}{\sqrt{x}}\right) d x = 20x^{\frac{9}{5}} - 6x^3 - 8x + 34\sqrt{x} + C$$

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

$$20x^{\frac{9}{5}} - 6x^3 - 8x + 34\sqrt{x} + C$$

Checking Our Work: The Beauty of Differentiation

One of the coolest things about calculus is that we can always check our work. How, you ask? By differentiating our result! If we differentiate the indefinite integral we found, we should get back the original function. This is because differentiation and integration are inverse operations.

Let's differentiate our answer and see if we arrive back at our starting point. Remember the power rule for differentiation: d/dx (x^n) = nx^(n-1). We'll apply this rule to each term in our result.

Differentiating 20x^(9/5)

$$\frac{d}{dx}(20x^{\frac{9}{5}}) = 20 \cdot \frac{9}{5} x^{\frac{9}{5}-1} = 36x^{\frac{4}{5}}$$

Differentiating -6x^3

$$\frac{d}{dx}(-6x^3) = -6 \cdot 3x^{3-1} = -18x^2$$

Differentiating -8x

$$\frac{d}{dx}(-8x) = -8$$

Differentiating 34√{x} (or 34x^(1/2))

$$\frac{d}{dx}(34x^{\frac{1}{2}}) = 34 \cdot \frac{1}{2} x^{\frac{1}{2}-1} = 17x^{-\frac{1}{2}} = \frac{17}{\sqrt{x}}$$

Differentiating the Constant C

$$\frac{d}{dx}(C) = 0$$

Putting It All Together (Again!)

Now, let's add up the derivatives of each term:

$$36x^{\frac{4}{5}} - 18x^2 - 8 + \frac{17}{\sqrt{x}} + 0 = 36 \sqrt[5]{x^4} - 18x^2 - 8 + \frac{17}{\sqrt{x}}$$

Boom! Look at that! We got back our original function. This confirms that our indefinite integral is correct. Pat yourselves on the back, guys – we nailed it!

Key Takeaways and Pro Tips

• Linearity is your friend: Always split integrals into smaller, manageable pieces. It simplifies the process immensely.
• Power rule is king: Master the power rule for integration and differentiation. It's the bread and butter of calculus.
• Constants matter: Don't forget the constant of integration, C. It represents the family of antiderivatives.
• Check your work: Differentiation is a fantastic way to verify your integration skills. It's like having a built-in answer key.
• Rewrite radicals as exponents: Converting radicals to fractional exponents makes applying the power rule much easier. For example, √x becomes x^(1/2) and ∛x becomes x^(1/3).
• Practice makes perfect: The more integrals you solve, the better you'll become. Don't be afraid to tackle challenging problems.

Real-World Applications: Why This Matters

Okay, so we've conquered this integral, but you might be wondering,