Problem 23
Question
Evaluate the following integrals. $$\int \frac{x^{2}+12 x-4}{x^{3}-4 x} d x$$
Step-by-Step Solution
Verified Answer
Answer: The integral is $$\ln |x| + \frac{2}{3} \ln |x^3 - 4| + C$$.
1Step 1: Rewrite the integrand using partial fraction decomposition
First, we can try to decompose the integrand into partial fractions. Let's rewrite the integrand in terms of partial fraction decomposition:
$$\frac{x^{2}+12x-4}{x^3-4x} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3-4}$$
To find the constants A, B, and C, we need to clear denominators:
x^2 + 12x - 4 = A(x^2)(x-4) + Bx(x-4) + Cx(x^2)
2Step 2: Solve for the constants A, B, and C
To solve for A, B, and C, we substitute convenient values for x to remove other terms:
- For A: let x = 0, then we have -4 = -4A ==> A = 1
- For B: differentiate both sides of the equation:
2x + 12 = 8A + (1-4x)B + x^2C
Substitute x = 0, we have 12 = 8(1) + 0B + 0C ==> B = 0
- For C: differentiate both sides of the equation one more time:
2 = -4B + 2xC
Substitute x = 0, we have 2 = -4(0) + C ==> C = 2
3Step 3: Rewrite integrand with found constants
Now that we have the values for A, B, and C, rewrite the integrand:
$$\frac{x^{2}+12x-4}{x^3-4x} = \frac{1}{x} + \frac{2}{x^3-4}$$
4Step 4: Integrate term by term
Next, integrate each term separately:
$$\int \frac{1}{x} dx = \ln |x| + C_1$$
$$\int \frac{2}{x^3-4} dx = \frac{2}{3} \int \frac{1}{u} du$$
(Here, we made a substitution u = x^3 - 4, so du = 3x^2 dx)
5Step 5: Combine the individual integrals
Now, we combine the two individual integrals to get the final result:
$$\int \frac{x^{2}+12 x-4}{x^{3}-4 x} d x = \ln |x| + \frac{2}{3} \ln |x^3 - 4| + C$$
Key Concepts
Integration TechniquesDefinite and Indefinite IntegralsSubstitution Method
Integration Techniques
Integration techniques are a set of strategic methods used to simplify and solve complex integrals. Often, the direct integration of complex expressions is not feasible. So, we use various techniques to break them down into simpler parts. Some common integration techniques include:
- Partial Fraction Decomposition: Breaking down a complex fraction into simpler fractions to make the integration process easier.
- Substitution Method: Changing the variable of integration to simplify the integral.
- Integration by Parts: A technique that removes the product by transforming it into a simpler integral.
Definite and Indefinite Integrals
When we talk about integrals, we generally refer to two types: definite and indefinite integrals. Definite integrals are used to calculate the area under a curve within a specific interval, while indefinite integrals are used to find the antiderivative of a function.
In our exercise, we dealt with an indefinite integral, which means the function was integrated without specifying any limits of integration. An indefinite integral is expressed as:\[ \int f(x) \, dx = F(x) + C \]where \(F(x)\) is the antiderivative of \(f(x)\) and \(C\) is the constant of integration.
Here, after decomposing the expression into simpler fractions, we found the antiderivatives of each term. We integrated term by term, resulting in: \[\ln |x| + \frac{2}{3} \ln |x^3 - 4| + C\]This encapsulates the complete antiderivative of the original function.
In our exercise, we dealt with an indefinite integral, which means the function was integrated without specifying any limits of integration. An indefinite integral is expressed as:\[ \int f(x) \, dx = F(x) + C \]where \(F(x)\) is the antiderivative of \(f(x)\) and \(C\) is the constant of integration.
Here, after decomposing the expression into simpler fractions, we found the antiderivatives of each term. We integrated term by term, resulting in: \[\ln |x| + \frac{2}{3} \ln |x^3 - 4| + C\]This encapsulates the complete antiderivative of the original function.
Substitution Method
The substitution method is a powerful technique used to simplify integrals, especially those involving composite functions or where a function can be represented in terms of another variable. This method involves introducing a new variable, \(u\), to replace an expression in the integrand.
Here's how substitution can simplify integration:
Through substitution, complex integrals can often be resolved into forms amenable to straightforward integration techniques, easing the overall integration process.
Here's how substitution can simplify integration:
- Identify a part of the integrand that can be replaced with the new variable \(u\).
- Compute the differential \(du\) in terms of \(dx\).
- Re-express the integral entirely in terms of \(u\) and \(du\).
Through substitution, complex integrals can often be resolved into forms amenable to straightforward integration techniques, easing the overall integration process.
Other exercises in this chapter
Problem 23
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