Problem 23
Question
Use the method of substitution to find each of the following indefinite integrals. $$ \int x\left(x^{2}+3\right)^{-12 / 7} d x $$
Step-by-Step Solution
Verified Answer
\(-\frac{7}{10}(x^2 + 3)^{-5/7} + C\).
1Step 1: Identify the Inner Function
The first step in substitution is identifying the part of the integrand to be substituted. Notice that the expression \(x^2 + 3\) can be substituted with a new variable \(u\). This makes it easier to integrate.
2Step 2: Define the Substitution
Set \(u = x^2 + 3\). Differentiate both sides to find the derivative: \(\frac{du}{dx} = 2x\), which implies \(du = 2x \, dx\).
3Step 3: Solve for the Differential and Substitute
We rearrange \(du = 2x \, dx\) to solve for \(x \, dx\): \(x \, dx = \frac{1}{2}du\). Substituting into the integral, we replace \((x^2 + 3)^{-12/7}\) with \(u^{-12/7}\), and \(x \, dx\) with \(\frac{1}{2}du\).
4Step 4: Integrate with Respect to \(u\)
The integral becomes \(\frac{1}{2} \int u^{-12/7} \, du\). The integral of \(u^n\), where \(n eq -1\), is \(\frac{u^{n+1}}{n+1}\). Here \(n = -12/7\), so \(\int u^{-12/7} \, du = \frac{u^{-5/7}}{-5/7}\).
5Step 5: Simplify the Expression
Simplify \(\frac{1}{2} \cdot \frac{u^{-5/7}}{-5/7}\) to get \(-\frac{7}{10}u^{-5/7} + C\) where \(C\) is the constant of integration.
6Step 6: Substitute Back the Original Variable
Replace \(u\) with \(x^2 + 3\) to express the integral in terms of \(x\). This gives the final answer: \(-\frac{7}{10}(x^2 + 3)^{-5/7} + C\).
Key Concepts
Indefinite IntegralsIntegration TechniquesCalculus Problem Solving
Indefinite Integrals
Indefinite integrals are a fundamental concept in calculus. They represent the most general form of antiderivatives and are used to find all possible functions whose derivative is a given function. An indefinite integral is denoted by the integral symbol \(\int\) followed by a function and \(dx\), which signifies integration concerning the variable \(x\). The result includes a constant of integration, \(C\), due to the property that derivatives of constants are zero.
Indefinite integrals answer the question: "What function differentiates to give this function?" For example, integrating \(\int 2x \, dx\) yields \(x^2 + C\), since the derivative of \(x^2\) is \(2x\). This process of integration connects closely with the concept of antiderivatives in calculus.
Indefinite integrals answer the question: "What function differentiates to give this function?" For example, integrating \(\int 2x \, dx\) yields \(x^2 + C\), since the derivative of \(x^2\) is \(2x\). This process of integration connects closely with the concept of antiderivatives in calculus.
Integration Techniques
Integration is more than reversing differentiation. It involves a set of techniques to solve different types of integrals. One powerful method is substitution, which simplifies complex integrals, making them easier to solve.
- **Substitution Method**: This involves substituting part of the integrand with a new variable, generally to simplify an expression. For instance, in the original exercise, the substitution was \(u = x^2 + 3\). After substitution, the integral reduces to a more manageable form.
- **Power Rule for Integration**: This is applicable when integrating functions in the form \(x^n\). The integral \(\int x^n \, dx\) is \(\frac{x^{n+1}}{n+1} + C\), provided \(n eq -1\).
Calculus Problem Solving
Solving calculus problems like the one given involves a systematic approach, using a variety of strategies and problem-solving skills. Here is how one might go about tackling such a problem:
- **Identify Parts**: First, look at the integral and identify if part of it can be simplified using substitution or other integration methods.
- **Correct Substitution**: Make sure that the derivative of the substituted part appears in the integrand, which allows the substitution to cancel out terms. In our case, substituting \(x^2 + 3\) with \(u\) and realizing \(du = 2x \, dx\) demonstrates this.
- **Integration by Substitution**: Integrate using the new variable, simplify, and then substitute back to the original variable.
Other exercises in this chapter
Problem 23
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