Chapter 1: Problem 43
In the following exercises, integrate using the indicated substitution. $$ \int \frac{1-x^{2}}{3 x-x^{3}} d x ; u=3 x-x^{3} $$
Short Answer
Expert verified
\(\frac{1}{3} \ln |3x - x^3| + C\).
Step by step solution
01
Assess the problem and substitution
The integral we need to evaluate is \(\int \frac{1-x^2}{3x-x^3} dx\). The substitution given is \(u=3x-x^3\). Our goal is to express the integral in terms of \(u\).
02
Differentiate the substitution
Differentiate the substitution \(u=3x-x^3\) to find \(du\) in terms of \(dx\). This gives us \(\frac{du}{dx} = 3 - 3x^2\), or equivalently, \(du = (3 - 3x^2) dx\).
03
Simplify the differential
Notice that \(du = 3(1-x^2) dx\). Divide both sides by 3 to isolate \((1-x^2)dx\): \((1-x^2)dx = \frac{1}{3}du\).
04
Substitute into the integral
Replace \(1-x^2)dx\) with \(\frac{1}{3}du\) in the integral. This converts the integral to \(\int \frac{1}{u} \cdot \frac{1}{3}du\), which simplifies to \(\frac{1}{3} \int \frac{1}{u} du\).
05
Integrate the simplified expression
The integral \(\int \frac{1}{u} du\) is a standard integral, which equals \(\ln |u| + C\). Thus, our expression \(\frac{1}{3} \int \frac{1}{u} du\) becomes \(\frac{1}{3} \ln |u| + C\).
06
Back-substitute to original variable
Recall that \(u = 3x - x^3\). Substitute this back into the integral result to express the answer in terms of \(x\): \(\frac{1}{3} \ln |3x - x^3| + C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Substitution
In calculus, the method of integration by substitution is a powerful tool to simplify complex integrals. It is akin to reversing the chain rule of differentiation. The main idea is to use a substitution to transform the integral into a simpler form. Let's consider the substitution \( u = 3x - x^3 \) from our example.
The differential \( du \) represents how \( u \) changes with respect to \( x \). By differentiating \( u = 3x - x^3 \), we find \( du = (3 - 3x^2) dx \). This differentiation allows us to express the integral's variables in terms of \( u \) and \( du \), effectively changing the variable of integration from \( x \) to \( u \).
Ultimately, this method transforms a complicated expression into a familiar and more manageable form, making the process of finding antiderivatives more straightforward.
The differential \( du \) represents how \( u \) changes with respect to \( x \). By differentiating \( u = 3x - x^3 \), we find \( du = (3 - 3x^2) dx \). This differentiation allows us to express the integral's variables in terms of \( u \) and \( du \), effectively changing the variable of integration from \( x \) to \( u \).
Ultimately, this method transforms a complicated expression into a familiar and more manageable form, making the process of finding antiderivatives more straightforward.
Definite Integral
The concept of definite integrals is fundamental in calculus, representing the area under a curve within specific bounds. However, in the exercise provided, we deal with an indefinite integral, which is slightly different. Indefinite integrals result in a family of functions and include a constant of integration, \( C \).
For definite integrals, the limits of integration are essential as they define the interval over which we compute the area. In contrast, for indefinite integrals, we're looking for the general expression representing a collection of all antiderivatives of a function.
For definite integrals, the limits of integration are essential as they define the interval over which we compute the area. In contrast, for indefinite integrals, we're looking for the general expression representing a collection of all antiderivatives of a function.
- Definite integrals have limits and yield a numerical value representing the area.
- Indefinite integrals do not have limits and yield a function that includes \( C \).
Differentiation
Differentiation is the process of finding the derivative of a function, which measures the rate of change. In our step-by-step solution, differentiation is used to find \( du \) from \( u = 3x - x^3 \).
The derivative of the expression \( 3x - x^3 \) is \( 3 - 3x^2 \). This result directly informs us about how \( u \) is changing with respect to \( x \). The relationship \( du = (3 - 3x^2) dx \) was then employed to manipulate the original integral into a simpler form by substitution.
This illustrates how closely linked differentiation and integration are. Often, solving integration problems requires a solid understanding of differentiation to reformulate the problem more conveniently.
The derivative of the expression \( 3x - x^3 \) is \( 3 - 3x^2 \). This result directly informs us about how \( u \) is changing with respect to \( x \). The relationship \( du = (3 - 3x^2) dx \) was then employed to manipulate the original integral into a simpler form by substitution.
This illustrates how closely linked differentiation and integration are. Often, solving integration problems requires a solid understanding of differentiation to reformulate the problem more conveniently.
Integral Calculus
Integral calculus is one half of the fundamental branches of calculus, with differentiation being the other half. It's concerned with accumulations, such as areas under curves, volumes, and other concepts that start with an elementary value and add up to a total.
Through integral calculus, we determine antiderivatives. An antiderivative of a function is essentially its reverse operation in differentiation. Our exercise required finding an antiderivative using the substitution method. This particular process allowed us to transform a complex integral into a simpler form, which could be easily integrated, illustrating the core purpose of integral calculus.
Understanding the integral calculus concepts like \( \int \frac{1}{u} \, du \) resulting in \( \ln |u| + C \) is critical. Such knowledge confirms how integral calculus not only facilitates finding sums and accumulations but also links closely with logarithmic functions.
Through integral calculus, we determine antiderivatives. An antiderivative of a function is essentially its reverse operation in differentiation. Our exercise required finding an antiderivative using the substitution method. This particular process allowed us to transform a complex integral into a simpler form, which could be easily integrated, illustrating the core purpose of integral calculus.
Understanding the integral calculus concepts like \( \int \frac{1}{u} \, du \) resulting in \( \ln |u| + C \) is critical. Such knowledge confirms how integral calculus not only facilitates finding sums and accumulations but also links closely with logarithmic functions.
