Chapter 5: Problem 66
Evaluate each definite integral. $$ \int_{1}^{2} \frac{(x+1)^{2}}{x^{2}} d x $$
Short Answer
Expert verified
The integral evaluates to \( \frac{3}{2} + 2 \ln(2) \).
Step by step solution
01
Simplify the Integrand
Rewrite the expression inside the integral by expanding the numerator before integrating. The original integral is: \( \int_{1}^{2} \frac{(x+1)^2}{x^2} \, dx \).Expanding \( (x + 1)^2 \), we get \( x^2 + 2x + 1 \).Thus, the integrand becomes:\[\int_{1}^{2} \left(\frac{x^2}{x^2} + \frac{2x}{x^2} + \frac{1}{x^2}\right) \, dx = \int_{1}^{2} \left(1 + \frac{2}{x} + \frac{1}{x^2}\right) \, dx\]
02
Integrate Term-by-Term
Now integrate each term separately over the interval \([1, 2]\):\[\int_{1}^{2} \left( 1 \right) \, dx + \int_{1}^{2} \left( \frac{2}{x} \right) \, dx + \int_{1}^{2} \left( \frac{1}{x^2} \right) \, dx\]- \( \int_{1}^{2} 1 \, dx = [x]_{1}^{2} \)- \( \int_{1}^{2} \frac{2}{x} \, dx = 2[\ln|x|]_{1}^{2} \)- \( \int_{1}^{2} \frac{1}{x^2} \, dx = [-\frac{1}{x}]_{1}^{2} \)
03
Evaluate the Integrals
Calculate the definite integrals obtained in Step 2:- For \( [x]_{1}^{2} \), calculate: \( 2 - 1 = 1 \)- For \( 2[\ln|x|]_{1}^{2} \), calculate: \( 2[\ln(2) - \ln(1)] = 2 \ln(2) \)- For \( [-\frac{1}{x}]_{1}^{2} \), calculate: \(-\frac{1}{2} - (-1) = \frac{1}{2} \)Add the results: \( 1 + 2 \ln(2) + \frac{1}{2} \)
04
Sum the Results
Combine all the evaluated parts:\[1 + 2 \ln(2) + \frac{1}{2} = \frac{3}{2} + 2 \ln(2)\]
05
Final Simplified Answer
Thus, the value of the definite integral is \( \frac{3}{2} + 2 \ln(2) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by parts is a technique used in calculus to integrate products of functions. It helps when the standard integration methods don’t work straightforwardly. You use the formula:
\[ \int u \, dv = uv - \int v \, du \]
- **Identify the parts:** Often, one function gets differentiated (typically a polynomial) and the other integrated (commonly exponential or trigonometric).- **Calculate components:** Choose which function in your integral should be "u" and which should be "dv."- **Substitution:** Substitute into the formula to simplify and solve the integral.
This method is particularly useful when dealing with products such as polynomial multiplied by a logarithmic or exponential function, though in this problem it wasn't needed since simplification was more appropriate.
\[ \int u \, dv = uv - \int v \, du \]
- **Identify the parts:** Often, one function gets differentiated (typically a polynomial) and the other integrated (commonly exponential or trigonometric).- **Calculate components:** Choose which function in your integral should be "u" and which should be "dv."- **Substitution:** Substitute into the formula to simplify and solve the integral.
This method is particularly useful when dealing with products such as polynomial multiplied by a logarithmic or exponential function, though in this problem it wasn't needed since simplification was more appropriate.
Polynomial Expansion
Polynomial expansion simplifies expressions with powers. It's crucial for converting a product of sums into a sum of products.
In the exercise, \( (x + 1)^2 \) is expanded to \( x^2 + 2x + 1 \). This transforms the integrand into terms that are easier to manipulate:- \( \int \frac{x^2}{x^2} \) becomes \( \int 1 \).- \( \int \frac{2x}{x^2} \) becomes \( \int \frac{2}{x} \).- \( \int \frac{1}{x^2} \) remains \( \int \frac{1}{x^2} \).Breaking it down this way simplifies the integration process. Each term fits a well-known integration formula, making computation straightforward. It's a powerful technique because it handles complex expressions by reducing them to simpler, manageable terms.
In the exercise, \( (x + 1)^2 \) is expanded to \( x^2 + 2x + 1 \). This transforms the integrand into terms that are easier to manipulate:- \( \int \frac{x^2}{x^2} \) becomes \( \int 1 \).- \( \int \frac{2x}{x^2} \) becomes \( \int \frac{2}{x} \).- \( \int \frac{1}{x^2} \) remains \( \int \frac{1}{x^2} \).Breaking it down this way simplifies the integration process. Each term fits a well-known integration formula, making computation straightforward. It's a powerful technique because it handles complex expressions by reducing them to simpler, manageable terms.
Logarithmic Integration
Logarithmic integration refers to integrating functions like \( \frac{1}{x} \), where the output involves logarithms.
For the term \( \int \frac{2}{x} \, dx \), the antiderivative is \( 2 \ln|x| \).Some essential points about logarithmic integration are:
For the term \( \int \frac{2}{x} \, dx \), the antiderivative is \( 2 \ln|x| \).Some essential points about logarithmic integration are:
- The integral of \( \frac{1}{x} \) is \( \ln|x| \), and the constant factor "2" remains outside the integral.
- Always consider the absolute value inside the logarithm to ensure it's defined across all real numbers where the function exists.
Simplification of Rational Functions
Simplification of rational functions is a vital skill for integrals, especially when dealing with fractions.
In our problem, initially, we have \( \frac{(x+1)^2}{x^2} \). Using polynomial expansion, it reduces to \( 1 + \frac{2}{x} + \frac{1}{x^2} \). This reformulation allows for easy integration.
In our problem, initially, we have \( \frac{(x+1)^2}{x^2} \). Using polynomial expansion, it reduces to \( 1 + \frac{2}{x} + \frac{1}{x^2} \). This reformulation allows for easy integration.
- The simplification helps in focusing on each separate part, especially when dealing with terms like \( \frac{2}{x} \) or \( \frac{1}{x^2} \).
- You ensure the integral goes faster since each piece uses standard formulas instead of struggling with the original fraction.