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Magazine Chapter 9: Problem 15
Evaluate the integral. $$\int \frac{x^{2}+2}{x+2} d x$$
Short Answer
Expert verified
\( \frac{x^2}{2} - 2x + 6 \ln |x+2| + C \)
Step by step solution
01
Polynomial Long Division
To simplify the integral, we first perform polynomial long division on the function \( \frac{x^2+2}{x+2} \). Divide \(x^2 + 0x + 2\) by \(x + 2\). The quotient is \(x - 2\) and the remainder is \(6\), so split the fraction as follows: \[ \int \left( x - 2 + \frac{6}{x+2} \right) \, dx \]
02
Separate the Integral
Separate the given integral into the sum of simpler integrals: \[ \int \left( x - 2 + \frac{6}{x+2} \right) \, dx = \int x \, dx - \int 2 \, dx + \int \frac{6}{x+2} \, dx \]
03
Integrate Each Term
Now, integrate each term separately:1. For \( \int x \, dx \), the integral is \( \frac{x^2}{2} \).2. For \( \int 2 \, dx \), the integral is \( 2x \).3. For \( \int \frac{6}{x+2} \, dx \), use the substitution \( u = x+2 \) followed by \( du = dx \). The integral becomes \( 6 \int \frac{1}{u} \, du \), which is \( 6 \ln |u| = 6 \ln |x+2| \).
04
Combine the Integrals
Combine all the integrated terms to form the final solution:\[ \frac{x^2}{2} - 2x + 6\ln|x+2| + C \] where \(C\) is the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Long Division
Polynomial long division helps simplify complex rational expressions into a series of simpler terms that can be integrated more easily. For our integral, we are dealing with the expression \( \frac{x^2+2}{x+2} \). To use polynomial long division:
- First, divide the leading term of the numerator \( x^2 \) by the leading term of the denominator \( x \), which gives \( x \).
- Multiply the entire divisor \( x + 2 \) by this result \( x \), yielding \( x^2 + 2x \).
- Subtract this from the original polynomial in the numerator \( x^2 + 2 \), leaving a remainder of \( -2x + 2 \).
- Repeat the process by dividing the new expression \(-2x + 2 \) by \( x + 2 \). This gives \( -2 \), and a new remainder of \( 6 \).
- This gives us the simplified expression \( x - 2 + \frac{6}{x+2} \), making it easier to integrate each term separately.
Integration Techniques
When faced with an integral like \( \int \left( x - 2 + \frac{6}{x+2} \right) \, dx \), using various integration techniques makes the process straightforward.
- For polynomial terms \( x \) and \( 2 \), we utilize basic integration rules: \( \int x \, dx = \frac{x^2}{2} \) and \( \int 2 \, dx = 2x \).
- The integral concerning \( \frac{6}{x+2} \) requires a more advanced method, often facilitated by substitution, which will be covered next.
- Combining these techniques helps solve integrals more efficiently and accurately, especially when properties of logarithms and polynomials are known.
Substitution Method
The substitution method is a critical strategy in calculus for simplifying integrals, especially those involving compositions of functions. For \( \int \frac{6}{x+2} \, dx \), the process is as follows:
- Identify a substitution that simplifies the fraction: let \( u = x + 2 \). This substitution changes the differential \( dx = du \).
- The integral becomes \( 6 \int \frac{1}{u} \, du \). This is a simpler integral that directly results in a logarithmic function.
- After integrating, the result is \( 6 \ln|u| + C \).
- Finally, substitute back \( u = x + 2 \) to obtain \( 6 \ln|x+2| + C \).
Logarithmic Integration
Logarithmic integration often arises when dealing with integrals of the form \( \int \frac{1}{x+a} \, dx \). For the term \( \int \frac{6}{x+2} \, dx \) in our example:
- Recognize that this is a standard form that integrates to a natural logarithm function.
- Perform the substitution, which simplifies the expression to \( 6 \int \frac{1}{u} \, du \).
- The integral \( \int \frac{1}{u} \, du \) results directly in \( \ln|u| \).
- Thus, the result of integrating \( 6 \int \frac{1}{x+2} \, dx \) yields \( 6 \ln|x+2| \), exhibiting the principles of logarithmic integration.
- Remember, this method can be applied to any function resembling \( \frac{1}{x+a} \), making it versatile and efficient for integration tasks.
