Solving Integrals: A Step-by-Step Guide

Hey Plastik Magazine readers! Let's dive into some serious math today, shall we? We're going to tackle an integral problem, and I'll break it down so even if you're not a math whiz, you can follow along. Our main goal is to evaluate the integral: $\int_0^1\left(\frac{12 x^2+24}{x^3+6 x+6}\right) d x$. Don't let the symbols scare you; we'll conquer this together! This guide will provide a clear, step-by-step approach to solve the problem, ensuring that you understand each step. We will break down the integral, making it easy to understand the fundamental concepts. We'll use a friendly tone, so it feels like a casual chat.

Understanding the Problem

First things first, what exactly are we dealing with? We have a definite integral. The limits of integration are from 0 to 1. Inside the integral, we have a fraction. The numerator is $12x^2 + 24$, and the denominator is $x^3 + 6x + 6$. Our task is to find the value of this integral. Before we jump into the calculation, let's take a deep breath. Integrals can seem intimidating, but they're just a way of finding the area under a curve. And in this case, we have to find the area under a curve represented by that fraction between x = 0 and x = 1. So, the question asks us to find the area under the curve of the function $f(x) = \frac{12 x^2+24}{x^3+6 x+6}$ from x = 0 to x = 1. The key to solving this type of integral lies in recognizing patterns and applying the right techniques.

Key Concepts and Techniques

The most useful technique here is u-substitution. It's a lifesaver when you see an integral that looks like a function and its derivative (or a multiple of it) are hanging out together. The basic idea of u-substitution is to simplify the integral by substituting a part of the expression with a new variable, 'u.' We also need to find the derivative of that part and then express the entire integral in terms of 'u' and 'du'. It's all about making the integral easier to solve. When we have a fraction with a more complex expression in the denominator, and the numerator looks like it could be related to the derivative of the denominator, u-substitution is usually the way to go. It is really important to keep a sharp eye out for these patterns. A good understanding of basic differentiation rules is also super important! Knowing how to differentiate polynomials and composite functions is vital. Also, some knowledge of logarithms is needed, especially how to integrate $\frac{1}{u}$. The integral of $\frac{1}{u}$ is $\ln|u|$. Knowing your basic integration rules will help you simplify your work and check if your answer is correct.

Step-by-Step Solution

Let's roll up our sleeves and solve the integral step-by-step. I'll make sure each step is clear as mud, so you won't get lost. Alright, let's go! Our starting point is the integral: $\int_0^1\left(\frac{12 x^2+24}{x^3+6 x+6}\right) d x$.

Step 1: Choosing Our 'u'

The goal here is to select a 'u' that will simplify the integral. Considering the form of the integral, let's choose $u = x^3 + 6x + 6$. The reason is because the derivative of $x^3 + 6x + 6$ is $3x^2 + 6$. We see $12x^2 + 24$ in the numerator, which is $4(3x^2 + 6)$, or 4 times the derivative of our 'u', so it's a good match.

Step 2: Finding 'du'

Now that we've chosen our 'u', we need to find its derivative, which is 'du'. So, if $u = x^3 + 6x + 6$, then $du = (3x^2 + 6) dx$. Notice that we need $12x^2 + 24$ in the numerator. We can rewrite the numerator as $4(3x^2 + 6)$. Thus, $12x^2 + 24 = 4(3x^2 + 6) = 4du$.

Step 3: Rewriting the Integral

We have $u = x^3 + 6x + 6$ and $du = (3x^2 + 6) dx$. Our integral becomes: $\int_0^1\left(\frac{12 x^2+24}{x^3+6 x+6}\right) d x = \int_0^1\frac{4du}{u}$.

Step 4: Integrating with Respect to 'u'

Now the integral is $\int_0^1\frac{4du}{u}$. The integral of $\frac{1}{u}$ is $\ln|u|$. So, $\int \frac{4}{u} du = 4\ln|u| + C$. We do not need to worry about the constant C here since we are dealing with a definite integral.

Step 5: Substituting Back

Replace 'u' with its original expression, $x^3 + 6x + 6$. So, we get $4\ln|x^3 + 6x + 6|$. Now that we've found the indefinite integral, we can find the definite integral with the limits 0 and 1.

Step 6: Applying the Limits of Integration

We need to evaluate the result from x = 0 to x = 1. So, we'll calculate $4\ln|x^3 + 6x + 6|$ at x = 1 and x = 0 and then subtract the two values.

At x = 1: $4\ln|1^3 + 6(1) + 6| = 4\ln|1 + 6 + 6| = 4\ln|13|$.

At x = 0: $4\ln|0^3 + 6(0) + 6| = 4\ln|0 + 0 + 6| = 4\ln|6|$.

Subtract the value at x = 0 from the value at x = 1: $4\ln(13) - 4\ln(6)$.

The Final Answer

And there you have it, guys! The correct answer is: $4\ln(13) - 4\ln(6)$. The answer corresponds to option B. I hope you found this guide helpful. Remember to practice regularly, and don't be afraid to break problems into smaller steps. Keep up the good work! And now you know how to conquer integrals!