Unlocking Integrals: Completing The Square Demystified!

Hey math enthusiasts! Today, we're diving deep into the world of calculus and tackling a classic problem: evaluating the indefinite integral of $\int \frac{x}{\sqrt{x^2-16x+39}} dx$. This integral might look a little intimidating at first glance, but fear not! We'll break it down step-by-step, using a powerful technique called completing the square to simplify things and make the integral much more manageable. Get ready to flex those math muscles and discover the beauty of integral calculus!

The Power of Completing the Square

So, what exactly does completing the square mean, and why is it so useful? In essence, completing the square is a clever algebraic manipulation that allows us to rewrite a quadratic expression (an expression of the form $ax^2 + bx + c$) into a more convenient form. Specifically, we aim to transform the expression into a perfect square trinomial plus a constant. This transformation is incredibly helpful when dealing with integrals involving quadratic expressions, especially those under square roots, as it allows us to simplify the integrand and often relate it to known integration formulas.

Let's take a closer look at the integral we're dealing with: $\int \frac{x}{\sqrt{x^2-16x+39}} dx$. Notice the quadratic expression $x^2 - 16x + 39$ under the square root. Our strategy will be to complete the square for this expression. This will transform the quadratic expression into a form like $(x - a)^2 + b$, where a and b are constants. This form is particularly useful because it allows us to simplify the integral and potentially use a substitution that makes it easier to solve. The core idea is to manipulate the quadratic to create a perfect square trinomial, which is a trinomial that can be factored into the square of a binomial, such as $(x - a)^2$. The constant term will be adjusted accordingly to maintain the equality of the expression.

To complete the square for $x^2 - 16x + 39$, we follow these steps: First, focus on the first two terms, $x^2 - 16x$. Take half of the coefficient of the x term (which is -16), square it ((-16/2)^2 = 64), and add and subtract it inside the expression. So, we have $x^2 - 16x + 64 - 64 + 39$. Now, we can rewrite the first three terms as a perfect square: $(x - 8)^2$. Simplifying the constants, we get $-64 + 39 = -25$. Therefore, $x^2 - 16x + 39$ becomes $(x - 8)^2 - 25$. This completed square form is exactly what we need to simplify our integral. The process of completing the square is a fundamental algebraic skill, not just for integration, but also for solving quadratic equations, analyzing conic sections, and more. Mastering this technique opens the door to solving a wide range of mathematical problems. Remember, practice makes perfect!

Applying Completing the Square to the Integral

Now that we've successfully completed the square, let's substitute this result back into our integral. We have: $\int \frac{x}{\sqrt{(x-8)^2-25}} dx$. The transformation significantly simplifies the expression under the square root. However, the x in the numerator is a bit of a problem. To deal with this, we'll use a clever trick involving a substitution. Our goal is to make the integrand look like a standard integral form that we know how to solve.

Let's consider a u-substitution. To get started, let $u = x - 8$. This implies that $x = u + 8$ and $du = dx$. Substituting these into the integral, we get: $\int \frac{u+8}{\sqrt{u^2-25}} du$

Now we can split this integral into two separate integrals: $\int \frac{u}{\sqrt{u^2-25}} du + \int \frac{8}{\sqrt{u^2-25}} du$. Let's solve them separately. The first integral, $\int \frac{u}{\sqrt{u^2-25}} du$, can be solved using another u-substitution. Let $v = u^2 - 25$, so $dv = 2u du$. Thus, $u du = \frac{1}{2} dv$. Substituting this into the first integral, we get: $\frac{1}{2} \int \frac{1}{\sqrt{v}} dv = \frac{1}{2} \int v^{-1/2} dv = \frac{1}{2} * 2v^{1/2} + C_1 = \sqrt{v} + C_1 = \sqrt{u^2-25} + C_1$.

The second integral, $\int \frac{8}{\sqrt{u^2-25}} du$, is a bit trickier, but it resembles a standard integral form: $\int \frac{1}{\sqrt{x^2-a^2}} dx = \ln|x + \sqrt{x^2-a^2}| + C$. In our case, $a = 5$. We can solve this integral using the formula. We can rewrite the second integral as: $8 \int \frac{1}{\sqrt{u^2-25}} du = 8 \ln|u + \sqrt{u^2-25}| + C_2$.

Putting It All Together

Now we can combine the results of both integrals, and substitute back $u = x-8$ into the result. Therefore: $\int \frac{x}{\sqrt{x^2-16x+39}} dx = \sqrt{x^2-16x+39} + 8 \ln|(x-8) + \sqrt{x^2-16x+39}| + C$.

So, we have successfully evaluated the indefinite integral! We started with an expression that looked quite daunting and, through the power of completing the square and strategic substitutions, managed to find its antiderivative. This example highlights the importance of recognizing patterns and utilizing appropriate techniques in integral calculus. The process may seem long, but each step is carefully designed to transform the integral into a more manageable form. Always remember to check your work. You can take the derivative of the result, and if you get back the original integrand, you know you have the correct answer. The process may seem long, but each step is carefully designed to transform the integral into a more manageable form.

Key Takeaways and Further Practice

• Completing the square is a versatile technique for simplifying quadratic expressions, especially within integrals. It's not just useful for integrals; you'll find it incredibly helpful in other areas of mathematics. Make sure you fully understand how to complete the square before moving on.
• Strategic substitutions are essential for solving many integrals. Choose your substitutions carefully to simplify the integral. Try to make the integrand resemble standard integral forms that you know. There are often multiple ways to approach an integral. Consider using different substitutions to solve it. This is where practice comes in!
• Understanding standard integral forms is crucial. Familiarize yourself with common integral formulas. This will save you time and make it easier to solve integrals. Many integral problems require a bit of creativity, so don't be afraid to experiment and try different approaches.

Now, armed with these techniques, try some practice problems. Here are a few to get you started:

1. $\int \frac{x}{\sqrt{x^2+4x+8}} dx$
2. $\int \frac{1}{\sqrt{x^2-6x+5}} dx$
3. $\int \frac{2x+3}{\sqrt{x^2-2x-3}} dx$

Remember to complete the square and use appropriate substitutions. Don't worry if you don't get the correct answers right away. The goal is to learn and improve. Check your answers, and don't be afraid to ask for help! The more you practice, the better you'll become at solving these types of problems. Calculus can be challenging, but it's also incredibly rewarding. Keep practicing, and you'll become a master of integrals in no time! Keep exploring, keep questioning, and keep having fun with math! Good luck, and happy integrating!