Solving The Integral: Step-by-Step Guide

Hey guys! Today, we're diving into a fascinating integral problem: $\displaystyle \int \frac{dx}{(x - 1) ^3 \sqrt{ {x}^{2} + 3x + 1 } }$. Integrals like this can seem daunting at first glance, but don't worry, we'll break it down together. Our main goal here is to explore the different techniques and strategies we can use to tackle this integral. We’ll go through a detailed, step-by-step solution that will not only help you understand this specific problem but also equip you with the tools to solve similar integrals in the future. So, let’s jump right into it and see how we can make this integral a piece of cake!

Understanding the Integral

Before we start crunching numbers and applying formulas, let's take a closer look at the integral we're dealing with: $\displaystyle \int \frac{dx}{(x - 1) ^3 \sqrt{ {x}^{2} + 3x + 1 } }$. The integrand, which is the expression inside the integral, is a rational function involving a square root. This means we'll likely need to use a combination of techniques, such as substitution, partial fractions, and possibly trigonometric substitution, to solve it.

Firstly, we need to identify the problematic parts. The term $(x - 1)^3$ in the denominator suggests that we might need to use partial fraction decomposition after a suitable substitution. The square root term, $\sqrt{ {x}^{2} + 3x + 1 }$, is another key element. Expressions involving square roots of quadratic polynomials often call for trigonometric substitutions. However, before we jump to trigonometric substitution, let’s see if a simpler substitution can simplify the square root term. Remember, the key to solving complex integrals is to break them down into smaller, more manageable parts. We want to transform the integral into a form that we can easily recognize and apply standard integration techniques to.

To successfully solve this integral, it's crucial to have a solid grasp of fundamental integration techniques. These include u-substitution, integration by parts, trigonometric substitutions, and partial fraction decomposition. We also need to be adept at algebraic manipulation and simplification. Each of these techniques serves as a tool in our problem-solving toolkit, and knowing when and how to apply them is essential. Think of it like having different tools in a toolbox – you need to know which tool is best suited for the job at hand. In the following sections, we will carefully select and apply the appropriate techniques to unravel this integral. So, let’s get our toolbox ready and start solving!

Initial Simplification and Substitution

Okay, let's get our hands dirty and start simplifying this integral! Our main goal in this section is to make the integral look a bit more friendly by using a clever substitution. We need to address both the $(x-1)^3$ term and the square root $\sqrt{x^2 + 3x + 1}$. A good starting point is often to try a substitution that simplifies the most complex part of the integrand. In this case, the square root is a prime candidate.

Let's consider the substitution: $u = x - 1$. This substitution is motivated by the presence of $(x-1)^3$ in the denominator, which will simplify nicely. We need to express everything in terms of $u$. If $u = x - 1$, then $x = u + 1$, and $dx = du$. Now, let’s substitute these into the square root term:

$$\begin{aligned} x^2 + 3x + 1 &= (u + 1)^2 + 3(u + 1) + 1 \\ &= u^2 + 2u + 1 + 3u + 3 + 1 \\ &= u^2 + 5u + 5 \end{aligned}$$

So, our integral now looks like this:

$$\int \frac{du}{u^3 \sqrt{u^2 + 5u + 5}}$$

This looks a bit better, doesn’t it? We’ve managed to simplify the $(x-1)^3$ term, but we still have a tricky square root. The next step is to deal with the square root term. To handle $\sqrt{u^2 + 5u + 5}$, we'll complete the square inside the square root. Completing the square will help us to identify a potential trigonometric substitution. Remember, the aim is to transform the quadratic expression inside the square root into a form that fits a standard trigonometric identity. By completing the square and performing a further substitution, we can hopefully simplify the integral into a more manageable form.

Completing the Square and Further Substitution

Now that we've made our initial substitution, let's tackle that pesky square root: $\sqrt{u^2 + 5u + 5}$. As we discussed, completing the square is the way to go here. This technique will transform the quadratic expression inside the square root into a form that is more amenable to trigonometric substitution. Let’s dive in!

To complete the square for $u^2 + 5u + 5$, we take half of the coefficient of the $u$ term (which is 5), square it (which gives us $\frac{25}{4}$), and add and subtract it within the expression:

$$\begin{aligned} u^2 + 5u + 5 &= u^2 + 5u + \frac{25}{4} - \frac{25}{4} + 5 \\ &= \left(u + \frac{5}{2}\right)^2 - \frac{25}{4} + \frac{20}{4} \\ &= \left(u + \frac{5}{2}\right)^2 - \frac{5}{4} \end{aligned}$$

So, our square root term becomes $\sqrt{\left(u + \frac{5}{2}\right)^2 - \frac{5}{4}}$. This looks much better! We now have a difference of squares inside the square root, which is a classic setup for a trigonometric substitution. Let's make another substitution to simplify this further. Let's set:

$$v = u + \frac{5}{2}$$

This implies that $u = v - \frac{5}{2}$ and $du = dv$. Substituting this into our integral, we get:

$$\int \frac{dv}{\left(v - \frac{5}{2}\right)^3 \sqrt{v^2 - \frac{5}{4}}}$$

Now, the integral is in a form that's screaming for a trigonometric substitution. The term $\sqrt{v^2 - \frac{5}{4}}$ suggests that we should use a secant substitution. Remember, the goal here is to eliminate the square root by using a trigonometric identity. By making the right trigonometric substitution, we can transform the integral into a trigonometric integral, which we can then solve using standard techniques.

Trigonometric Substitution

Alright, guys, we've made some great progress! We've simplified the integral quite a bit, and now it's time for the main event: the trigonometric substitution. As we discussed, the form of the integral, specifically the $\sqrt{v^2 - \frac{5}{4}}$ term, strongly suggests a secant substitution. Let’s see how this works.

We'll use the substitution:

$$v = \frac{\sqrt{5}}{2} \sec(\theta)$$

This substitution is based on the trigonometric identity $\sec^2(\theta) - 1 = \tan^2(\theta)$. When we substitute this into the square root, we should get a nice simplification. First, let's find $dv$:

$$dv = \frac{\sqrt{5}}{2} \sec(\theta) \tan(\theta) \, d\theta$$

Now, let's substitute $v$ into the square root term:

$$\begin{aligned} \sqrt{v^2 - \frac{5}{4}} &= \sqrt{\frac{5}{4} \sec^2(\theta) - \frac{5}{4}} \\ &= \sqrt{\frac{5}{4}(\sec^2(\theta) - 1)} \\ &= \sqrt{\frac{5}{4} \tan^2(\theta)} \\ &= \frac{\sqrt{5}}{2} |\tan(\theta)| \end{aligned}$$

We'll assume that $\tan(\theta)$ is positive for the range of $\theta$ we're considering, so we can drop the absolute value signs. Now, let's substitute $v$ and $dv$ into the integral:

$$\int \frac{\frac{\sqrt{5}}{2} \sec(\theta) \tan(\theta) \, d\theta}{\left(\frac{\sqrt{5}}{2} \sec(\theta) - \frac{5}{2}\right)^3 \frac{\sqrt{5}}{2} \tan(\theta)}$$

We can simplify this by canceling out the $\frac{\sqrt{5}}{2} \tan(\theta)$ terms:

$$\int \frac{d\theta}{\left(\frac{\sqrt{5}}{2} \sec(\theta) - \frac{5}{2}\right)^3}$$

This simplifies further to:

$$\int \frac{d\theta}{\frac{5\sqrt{5}}{8} (\sec(\theta) - \sqrt{5})^3}$$

Which can be written as:

$$\frac{8}{5\sqrt{5}} \int \frac{d\theta}{(\sec(\theta) - \sqrt{5})^3}$$

This trigonometric integral looks complex, but we've made significant progress by eliminating the square root. The next step involves using trigonometric identities and potentially further substitutions to simplify the integrand. By carefully applying trigonometric identities and algebraic manipulations, we can transform this complex trigonometric integral into a form that we can integrate directly.

Simplifying the Trigonometric Integral

Okay, team, we've arrived at a crucial stage. We've successfully performed the trigonometric substitution, and now we're faced with the integral:$\frac{8}{5\sqrt{5}} \int \frac{d\theta}{(\sec(\theta) - \sqrt{5})^3}$ This looks pretty intimidating, but don't worry, we can handle it. The key here is to manipulate the integrand using trigonometric identities to make it more manageable. A common strategy when dealing with powers of trigonometric functions is to express everything in terms of sines and cosines. Let's do that:

Recall that $\sec(\theta) = \frac{1}{\cos(\theta)}$. Substituting this into our integral, we get:

$$\frac{8}{5\sqrt{5}} \int \frac{d\theta}{\left(\frac{1}{\cos(\theta)} - \sqrt{5}\right)^3}$$

Now, let's simplify the denominator by finding a common denominator:

$$\frac{8}{5\sqrt{5}} \int \frac{d\theta}{\left(\frac{1 - \sqrt{5}\cos(\theta)}{\cos(\theta)}\right)^3}$$

This can be rewritten as:

$$\frac{8}{5\sqrt{5}} \int \frac{\cos^3(\theta)}{(1 - \sqrt{5}\cos(\theta))^3} \, d\theta$$

This integral is still challenging, but it's in a form where we can see a potential strategy. We have a power of cosine in the numerator and a power of a linear expression involving cosine in the denominator. One approach here is to try a substitution that simplifies the denominator. Let's try the substitution:

$$w = 1 - \sqrt{5}\cos(\theta)$$

Then, $dw = \sqrt{5}\sin(\theta) \, d\theta$. This substitution will simplify the denominator, but we'll need to express $\cos(\theta)$ and $\cos^3(\theta)$ in terms of $w$. From our substitution, we have:

$$\cos(\theta) = \frac{1 - w}{\sqrt{5}}$$

Now, we need to find an expression for $\cos^3(\theta)$. We can cube both sides of the above equation:

$$\cos^3(\theta) = \left(\frac{1 - w}{\sqrt{5}}\right)^3 = \frac{(1 - w)^3}{5\sqrt{5}}$$

Substituting these expressions into the integral, we get:

$$\frac{8}{5\sqrt{5}} \int \frac{\frac{(1 - w)^3}{5\sqrt{5}}}{w^3} \cdot \frac{dw}{\sqrt{5}\sin(\theta)}$$

This looks even more complicated, but we're getting closer! We still have a $\sin(\theta)$ term in the denominator, which we need to express in terms of $w$. We can use the identity $\sin^2(\theta) = 1 - \cos^2(\theta)$ to help us with this. By continuing to use substitutions and trigonometric identities, we can systematically simplify the integral until we reach a form that we can integrate directly.

Handling the Remaining Trigonometric Terms

Okay, guys, let's keep pushing forward! We've made some significant progress, and we're now dealing with the integral:

$$\frac{8}{5\sqrt{5}} \int \frac{\frac{(1 - w)^3}{5\sqrt{5}}}{w^3} \cdot \frac{dw}{\sqrt{5}\sin(\theta)}$$

The tricky part here is the $\sin(\theta)$ term in the denominator. As we discussed, we can use the identity $\sin^2(\theta) = 1 - \cos^2(\theta)$ to express $\sin(\theta)$ in terms of $w$. Let's do that:

We know that $\cos(\theta) = \frac{1 - w}{\sqrt{5}}$, so

$$\begin{aligned} \sin^2(\theta) &= 1 - \cos^2(\theta) \\ &= 1 - \left(\frac{1 - w}{\sqrt{5}}\right)^2 \\ &= 1 - \frac{(1 - w)^2}{5} \\ &= \frac{5 - (1 - 2w + w^2)}{5} \\ &= \frac{4 + 2w - w^2}{5} \end{aligned}$$

Thus,

$$\sin(\theta) = \pm \sqrt{\frac{4 + 2w - w^2}{5}}$$

We'll take the positive square root for simplicity. Now we can substitute this expression for $\sin(\theta)$ into our integral:

$$\frac{8}{5\sqrt{5}} \int \frac{\frac{(1 - w)^3}{5\sqrt{5}}}{w^3} \cdot \frac{dw}{\sqrt{5} \sqrt{\frac{4 + 2w - w^2}{5}}}$$

Let's simplify this a bit. We can cancel out a $\sqrt{5}$ from the denominator:

$$\frac{8}{25} \int \frac{(1 - w)^3}{w^3} \cdot \frac{dw}{\sqrt{4 + 2w - w^2}}$$

This integral is still quite challenging, but we've managed to express it entirely in terms of $w$. The integrand now involves a rational function multiplied by the reciprocal of a square root. At this point, we might consider using partial fraction decomposition on the rational function part of the integrand. By breaking down the rational function into simpler fractions, we can potentially make the integral more manageable and easier to integrate.

Partial Fraction Decomposition and Final Integration Steps

Alright, we're in the home stretch now! We've simplified the integral to:

$$\frac{8}{25} \int \frac{(1 - w)^3}{w^3 \sqrt{4 + 2w - w^2}} \, dw$$

As we discussed, let's use partial fraction decomposition on the rational function part of the integrand. First, let's expand $(1 - w)^3$:

$$(1 - w)^3 = 1 - 3w + 3w^2 - w^3$$

So, our integral becomes:

$$\frac{8}{25} \int \frac{1 - 3w + 3w^2 - w^3}{w^3 \sqrt{4 + 2w - w^2}} \, dw$$

We can rewrite the rational function as a sum of fractions:

$$\frac{1 - 3w + 3w^2 - w^3}{w^3} = \frac{1}{w^3} - \frac{3}{w^2} + \frac{3}{w} - 1$$

Thus, our integral becomes:

$$\frac{8}{25} \int \left(\frac{1}{w^3} - \frac{3}{w^2} + \frac{3}{w} - 1\right) \frac{1}{\sqrt{4 + 2w - w^2}} \, dw$$

Now, we have a sum of integrals:

$$\frac{8}{25} \left[ \int \frac{dw}{w^3 \sqrt{4 + 2w - w^2}} - 3 \int \frac{dw}{w^2 \sqrt{4 + 2w - w^2}} + 3 \int \frac{dw}{w \sqrt{4 + 2w - w^2}} - \int \frac{dw}{\sqrt{4 + 2w - w^2}} \right]$$

Each of these integrals can be tackled using various techniques, such as further substitutions or integral tables. The last integral, $\int \frac{dw}{\sqrt{4 + 2w - w^2}}$, can be solved by completing the square inside the square root and using a trigonometric substitution. The other integrals might require more advanced techniques or the use of a computer algebra system.

At this point, we've broken down the original integral into a sum of simpler integrals. While each of these integrals might still require some work to solve, we've made significant progress in simplifying the problem. Remember, the key to solving complex integrals is to break them down into manageable parts and apply the appropriate techniques. By systematically applying the techniques of substitution, completing the square, trigonometric substitution, and partial fraction decomposition, we can solve even the most challenging integrals.

Final Thoughts

Alright, guys, we've reached the end of our journey through this complex integral! We started with a daunting expression and, through a series of clever substitutions, trigonometric identities, and partial fraction decomposition, we've managed to break it down into a sum of simpler integrals. While the final steps might still require some effort, we've laid the groundwork for a solution.

Solving integrals like this is a testament to the power of mathematical techniques and problem-solving strategies. It's like piecing together a puzzle – each technique is a piece, and when you fit them together correctly, you get a beautiful solution. The journey through this integral highlights the importance of persistence, attention to detail, and a solid understanding of fundamental calculus concepts.

I hope this step-by-step guide has been helpful and has given you a better understanding of how to tackle complex integrals. Remember, practice makes perfect, so keep exploring and keep solving! Happy integrating, everyone!