Evaluating ∫ (16x^3 + 64x^2 + X + 2) / (16x^2 + 1) Dx With Long Division

Hey guys! Today, we're diving into the exciting world of calculus to tackle an integral that looks a bit intimidating at first glance. We've got a rational function here, where the degree of the numerator is higher than the degree of the denominator. Don't worry, we'll break it down step-by-step. The integral we're going to evaluate is:

$$\int \frac{16 x^3+64 x^2+x+2}{16 x^2+1} d x$$

The key to solving this type of integral is polynomial long division. This technique allows us to simplify the integrand into a form that's easier to integrate. Let's get started!

Polynomial Long Division

When we encounter an integral where the degree of the numerator is greater than or equal to the degree of the denominator, polynomial long division is our best friend. Think of it like long division with numbers, but with polynomials instead. The main goal here is to rewrite the integrand as the sum of a quotient and a remainder divided by the original denominator. This makes the integration process much more manageable.

So, let's dive into the long division process with our given integrand, (16x^3 + 64x^2 + x + 2) / (16x^2 + 1). We'll set it up just like regular long division. First, we divide the leading term of the numerator (16x^3) by the leading term of the denominator (16x^2), which gives us x. This is the first term of our quotient. Next, we multiply the entire denominator (16x^2 + 1) by x, resulting in 16x^3 + x. We then subtract this from the numerator. This subtraction cancels out the 16x^3 term and the x term, leaving us with 64x^2 + 2. Now, we bring down the next term, which isn't necessary in this case since we already have all the terms accounted for. We repeat the process with the new polynomial, 64x^2 + 2. Divide the leading term (64x^2) by the leading term of the denominator (16x^2), which gives us 4. This is the next term in our quotient. Multiply the entire denominator (16x^2 + 1) by 4, resulting in 64x^2 + 4. Subtract this from 64x^2 + 2, and we're left with a remainder of -2. Now we can express our original fraction as the sum of the quotient and the remainder divided by the divisor. In this case, that's x + 4, which is our quotient, and -2, which is our remainder, all over the original denominator, 16x^2 + 1. This transformation is crucial because it simplifies the integral into manageable parts. The quotient, x + 4, is a simple polynomial that we can easily integrate. The remainder part, -2 / (16x^2 + 1), might look tricky, but it's a form that we can tackle using techniques like u-substitution or recognizing it as a form related to the arctangent function. By performing polynomial long division, we've successfully transformed a complex rational function into a sum of simpler terms that are far easier to integrate. This is a common and powerful strategy in calculus, especially when dealing with rational functions where the degree of the numerator is greater than or equal to the degree of the denominator. It’s a bit like breaking a large, challenging problem into smaller, more solvable pieces, making the overall task much less daunting. Remember, this technique isn't just about finding the answer; it's about understanding the structure of the problem and applying the right tools to simplify it. So, embrace the power of polynomial long division, and you'll find that many seemingly complicated integrals become much more approachable.

After performing the long division, we find that:

$$\frac{16 x^3+64 x^2+x+2}{16 x^2+1} = x + 4 + \frac{-2}{16 x^2+1}$$

Now, our integral becomes:

$$\int \left(x + 4 - \frac{2}{16 x^2+1}\right) d x$$

Integrating the Quotient and Remainder

Now that we've used polynomial long division to simplify our integral, we're ready to integrate each term separately. This is where the magic of calculus really shines, as we break down a complex problem into smaller, more manageable parts. Remember, the power of integration lies in its ability to find the antiderivative of a function, essentially reversing the process of differentiation. When we have a sum or difference of terms inside an integral, we can integrate each term individually, making the whole process much more straightforward.

First, let's tackle the integral of x. This is a basic power rule integration, where we increase the exponent by one and divide by the new exponent. So, the integral of x becomes (x^2) / 2. Next up, we have the integral of 4. Since 4 is a constant, its integral is simply 4x. Now, let's address the more interesting part: the integral of -2 / (16x^2 + 1). This might look intimidating at first, but it's a classic form that we can solve using a bit of algebraic manipulation and our knowledge of inverse trigonometric functions. The key here is to recognize that the denominator, 16x^2 + 1, is similar to the form a^2 * u^2 + 1, which appears in the derivative of the arctangent function. To make our integral fit this form perfectly, we need to factor out a 16 from the denominator, which gives us 16(x^2 + 1/16). This allows us to rewrite the integral as -2 / (16(x^2 + 1/16)). Now, we can bring the 16 outside the integral, making it -1/8 * integral of 1 / (x^2 + (1/4)^2) dx. This form is exactly the arctangent integral, where the integral of 1 / (u^2 + a^2) du is (1/a) * arctan(u/a) + C. In our case, a is 1/4 and u is x. Applying this formula, we find that the integral of 1 / (x^2 + (1/4)^2) is 4 * arctan(4x). Multiplying by the -1/8 factor we had outside the integral, we get -1/2 * arctan(4x). Putting it all together, the integral of our simplified expression, x + 4 - 2 / (16x^2 + 1), is (x^2) / 2 + 4x - 1/2 * arctan(4x) + C, where C is the constant of integration. Remember, we always add C when finding indefinite integrals because the derivative of a constant is zero, meaning there are infinitely many antiderivatives that differ only by a constant. Integrating each term separately is not just a technique; it's a way of thinking about calculus problems. By breaking down complex integrals into simpler parts, we can apply known rules and formulas to each part, making the overall problem much more approachable. This approach not only helps in solving integrals but also builds a deeper understanding of the underlying concepts of calculus. So, next time you encounter a daunting integral, remember the power of separation and conquer each term individually!

We can now integrate term by term:

$$\int x \, dx = \frac{x^2}{2} + C_1$$

$$\int 4 \, dx = 4x + C_2$$

$$\int \frac{-2}{16 x^2+1} \, dx = -2 \int \frac{1}{16 x^2+1} \, dx$$

For the last integral, we'll use a u-substitution. Let $u = 4x$, so $du = 4 dx$ and $dx = \frac{du}{4}$. Thus,

$$-2 \int \frac{1}{16 x^2+1} \, dx = -2 \int \frac{1}{u^2+1} \cdot \frac{du}{4} = -\frac{1}{2} \int \frac{1}{u^2+1} \, du$$

We know that $\int \frac{1}{u^2+1} \, du = \arctan(u) + C_3$, so

$$- \frac{1}{2} \int \frac{1}{u^2+1} \, du = -\frac{1}{2} \arctan(4x) + C_3$$

Final Solution

Alright guys, we've done the heavy lifting – the long division, the term-by-term integration, and the tricky u-substitution. Now, it's time to gather all the pieces and assemble our final solution. This is where we see how all the individual steps come together to give us the complete antiderivative of the original function. Think of it like putting together a puzzle; each piece (each integral we solved) fits perfectly to reveal the final image (the solution). We started with a complex rational function and, through careful application of calculus techniques, transformed it into something much simpler and more elegant. This final step is not just about writing down the answer; it's about appreciating the journey and the process that got us here.

So, let's revisit what we've found so far. We integrated x and got (x^2) / 2. We integrated the constant 4 and obtained 4x. And then, we tackled the tricky fraction -2 / (16x^2 + 1), using a u-substitution to transform it into a standard arctangent integral, which gave us -1/2 * arctan(4x). Now, we simply add these results together, being sure to include our constant of integration, C. This constant is crucial because it acknowledges that the antiderivative is not unique; there are infinitely many functions that have the same derivative, differing only by a constant value. Therefore, our final solution is (x^2) / 2 + 4x - 1/2 * arctan(4x) + C. This is the antiderivative of our original function, and it represents a family of functions, each differing by a constant, whose derivatives are equal to the integrand we started with. But hold on, before we declare victory, let's take a moment to appreciate the power of calculus. We started with a seemingly daunting integral, a complex fraction that looked intimidating. But by applying the right techniques – polynomial long division to simplify the fraction, term-by-term integration to break down the problem, and u-substitution to handle the tricky part – we were able to find the solution. This process highlights the beauty of calculus: its ability to transform complex problems into manageable steps and to reveal the underlying structure of mathematical relationships. So, let’s add all those results together:

$$\int \frac{16 x^3+64 x^2+x+2}{16 x^2+1} d x = \frac{x^2}{2} + 4x - \frac{1}{2} \arctan(4x) + C$$

Where $C = C_1 + C_2 + C_3$ is the constant of integration. And there you have it! We've successfully evaluated the integral. Remember, the key is to break down complex problems into smaller, manageable steps. Keep practicing, and you'll become a master of integration in no time!