- Partial fraction decomposition is the process of expressing a rational function as the sum of simpler fractions.
- It is particularly useful when integrating rational functions.
- In this lecture, we will look at examples of different forms of partial fractions and how to integrate them.
- If the denominator of a rational function has distinct linear factors, then we can write the partial fraction decomposition as follows:
- Example: Consider the function
- ( \frac{1}{x(x + 1)(x - 2)} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 2} )
- We can find the values of ( A ), ( B ), and ( C ) through various methods, such as comparing coefficients or solving a system of equations.
- Given the function ( \frac{2x + 3}{x(x + 2)(x - 1)} )
- We can write it in partial fraction form as:
- ( \frac{2x + 3}{x(x + 2)(x - 1)} = \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 1} )
- To find the values of ( A ), ( B ), and ( C ), we can perform a common denominator and equate coefficients.
- If the denominator of a rational function has repeated linear factors, then the partial fraction decomposition will have additional terms.
- Example: Consider the function
- ( \frac{4x + 3}{(x - 1)^3} )
- The partial fraction decomposition of this function will have the following form:
- ( \frac{4x + 3}{(x - 1)^3} = \frac{A}{(x - 1)} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} )
- Again, we can find the values of ( A ), ( B ), and ( C ) using various methods.
- Given the function ( \frac{3x^2 + 5x + 7}{(x + 1)^2} )
- The partial fraction decomposition will be:
- ( \frac{3x^2 + 5x + 7}{(x + 1)^2} = \frac{A}{(x + 1)} + \frac{B}{(x + 1)^2} )
- To find the values of ( A ) and ( B ), we can equate coefficients and simplify the equations.
- If the denominator of a rational function has quadratic factors, then the partial fraction decomposition will involve constants and linear terms.
- Example: Consider the function
- ( \frac{2x + 1}{x^2 + 3x + 2} )
- The partial fraction decomposition of this function will be:
- ( \frac{2x + 1}{x^2 + 3x + 2} = \frac{A}{x + 1} + \frac{B}{x + 2} )
- By equating coefficients and solving equations, we can determine the values of ( A ) and ( B ).
- Given the function ( \frac{4x^2 + 5x - 1}{x^2 - 2x + 1} )
- The partial fraction decomposition will be:
- ( \frac{4x^2 + 5x - 1}{x^2 - 2x + 1} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} )
- We can find the values of ( A ) and ( B ) by equating coefficients and solving the resulting equations.
- If the denominator of a rational function has irreducible quadratic factors, then the partial fraction decomposition will involve linear terms with quadratic denominators.
- Example: Consider the function
- ( \frac{3x + 2}{x^2 + 4x + 5} )
- The partial fraction decomposition of this function will be:
- ( \frac{3x + 2}{x^2 + 4x + 5} = \frac{Ax + B}{x^2 + 4x + 5} )
- Here, ( Ax + B ) represents the linear term that cannot be further simplified.
- Given the function ( \frac{x^2 - x + 2}{x^2 + 4x + 5} )
- The partial fraction decomposition will be:
- ( \frac{x^2 - x + 2}{x^2 + 4x + 5} = \frac{Ax + B}{x^2 + 4x + 5} )
- By comparing coefficients and solving the resulting equations, we can determine the values of ( A ) and ( B ).
Recap
- Today, we discussed examples of different forms of partial fractions and how to integrate them.
- We looked at partial fraction decomposition for functions with distinct linear factors, repeated linear factors, quadratic factors, and irreducible quadratic factors.
- Partial fraction decomposition is a crucial technique when integrating rational functions.
- Practice solving more examples to strengthen your understanding.
Review: Indefinite Integral
- The indefinite integral of a function ( f(x) ) is denoted as ( \int f(x) , dx ) and represents the antiderivative of ( f(x) ).
- It is the reverse process of differentiation.
- The indefinite integral of a continuous function ( f(x) ) yields a family of functions with different constants of integration.
Applying Partial Fraction Decomposition to Indefinite Integration
- When integrating rational functions, partial fraction decomposition can make the integration process easier.
- It allows us to break down a complex fraction into simpler, more manageable components.
- By doing this, we can integrate each term of the partial fraction separately.
Steps for Integrating Partial Fractions
- Perform partial fraction decomposition to express the rational function as a sum of simpler fractions.
- Rewrite each fraction as a power of ( x ).
- Integrate each fraction using the power rule for integration.
- If necessary, perform additional simplification to express the final answer.
Example: Integrating a Function with Distinct Linear Factors
- Given the function ( \frac{2x + 3}{x(x + 2)(x - 1)} )
- We previously found its partial fraction decomposition as:
- ( \frac{2x + 3}{x(x + 2)(x - 1)} = \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 1} )
- Now, let’s integrate each term separately.
- ( \int \frac{2x + 3}{x(x + 2)(x - 1)} , dx = \int \frac{A}{x} , dx + \int \frac{B}{x + 2} , dx + \int \frac{C}{x - 1} , dx )
Example (continued): Integrating a Function with Distinct Linear Factors
- Integrating each term separately:
- ( \int \frac{A}{x} , dx = A \ln|x| + C_1 )
- ( \int \frac{B}{x + 2} , dx = B \ln|x + 2| + C_2 )
- ( \int \frac{C}{x - 1} , dx = C \ln|x - 1| + C_3 )
- Where ( C_1, C_2, C_3 ) are constants of integration.
- The final answer will be the sum of these integrals.
Example: Integrating a Function with Repeated Linear Factors
- Given the function ( \frac{3x^2 + 5x + 7}{(x + 1)^2} )
- The partial fraction decomposition is:
- ( \frac{3x^2 + 5x + 7}{(x + 1)^2} = \frac{A}{(x + 1)} + \frac{B}{(x + 1)^2} )
- Now, let’s integrate each term individually.
- ( \int \frac{3x^2 + 5x + 7}{(x + 1)^2} , dx = \int \frac{A}{(x + 1)} , dx + \int \frac{B}{(x + 1)^2} , dx )
Example (continued): Integrating a Function with Repeated Linear Factors
- Integrating each term separately:
- ( \int \frac{A}{(x + 1)} , dx = A \ln|x + 1| + C_1 )
- ( \int \frac{B}{(x + 1)^2} , dx = -B \frac{1}{x + 1} + C_2 )
- Where ( C_1 ) and ( C_2 ) are constants of integration.
- The final answer will be the sum of these integrals.
Example: Integrating a Function with Quadratic Factors
- Given the function ( \frac{2x + 1}{x^2 + 3x + 2} )
- The partial fraction decomposition is:
- ( \frac{2x + 1}{x^2 + 3x + 2} = \frac{A}{x + 1} + \frac{B}{x + 2} )
- Now, let’s integrate each term separately.
- ( \int \frac{2x + 1}{x^2 + 3x + 2} , dx = \int \frac{A}{x + 1} , dx + \int \frac{B}{x + 2} , dx )
Example (continued): Integrating a Function with Quadratic Factors
- Integrating each term individually:
- ( \int \frac{A}{x + 1} , dx = A \ln|x + 1| + C_1 )
- ( \int \frac{B}{x + 2} , dx = B \ln|x + 2| + C_2 )
- Where ( C_1 ) and ( C_2 ) are constants of integration.
- The final answer will be the sum of these integrals.
Recap
- In this lecture, we looked at various examples of integrating partial fractions.
- We covered both distinct and repeated linear factors, as well as quadratic factors in the partial fraction decomposition.
- Remember to integrate each term separately by applying the power rule for integration.
- Practicing more examples will help you gain confidence in this topic.
Slide 21:
- Form 5: Mixed Factors in the Denominator
- When the denominator of a rational function has a combination of distinct linear factors, repeated linear factors, and/or quadratic factors, the partial fraction decomposition will involve a combination of linear terms and quadratic terms.
- Example: ( \frac{1}{x(x + 1)(x - 1)^2(x^2 + 1)} )
- The partial fraction decomposition will have terms like ( \frac{A}{x} ), ( \frac{B}{x + 1} ), ( \frac{C}{(x - 1)} ), ( \frac{D}{(x - 1)^2} ), and ( \frac{Ex + F}{x^2 + 1} )
- Solving for the values of ( A ), ( B ), ( C ), ( D ), ( E ), and ( F ) will require a system of equations to be solved.
Slide 22:
- Example of Form 5: Mixed Factors
- Given the function ( \frac{3x + 1}{x(x + 1)(x - 1)^2(x^2 + 1)} )
- The partial fraction decomposition will involve terms like ( \frac{A}{x} ), ( \frac{B}{x + 1} ), ( \frac{C}{(x - 1)} ), ( \frac{D}{(x - 1)^2} ), and ( \frac{Ex + F}{x^2 + 1} )
- By solving a system of equations, we can determine the values of ( A ), ( B ), ( C ), ( D ), ( E ), and ( F ) to complete the partial fraction decomposition.
Slide 23:
- Choosing the Appropriate Method for Finding Coefficients
- There are multiple methods that can be used to find the coefficients in the partial fraction decomposition.
- For simple cases, comparing coefficients or equating the numerators may be sufficient.
- For more complex cases, solving a system of equations or using algebraic manipulation may be necessary.
- Choose the method that suits the given rational function and allows for easier solving.
Slide 24:
- Review: Power Rule for Integration
- The power rule states that ( \int x^n , dx = \frac{x^{n+1}}{n+1} + C )
- This rule applies when integrating terms with a variable raised to a power.
- Here, ( n ) is a real number that is not equal to -1, and ( C ) represents the constant of integration.
Slide 25:
- Review: Natural Logarithm Function
- The natural logarithm function, denoted as ( \ln(x) ), is the inverse of the exponential function.
- It is defined only for positive real numbers.
- Integrating the reciprocal of a function gives the natural logarithm of the absolute value of the function.
- Example: ( \int \frac{1}{x} , dx = \ln|x| + C )
Slide 26:
- Integrating Trigonometric Functions
- Trigonometric functions like sine, cosine, and tangent can also be integrated using specific rules.
- These rules involve trigonometric identities and substitution techniques.
- The integrals of trigonometric functions often give logarithmic or inverse trigonometric functions as solutions.
Slide 27:
- Example: Integrating Trigonometric Functions
- Given the function ( \int \sin(x) , dx )
- The integral of the sine function is equal to the negative cosine function, plus a constant of integration.
- ( \int \sin(x) , dx = -\cos(x) + C )
Slide 28:
- Example: Integrating Exponential Functions
- Exponential functions like ( e^x ) can be integrated using a straightforward rule.
- The integral of ( e^x ) is equal to itself, plus a constant of integration.
- ( \int e^x , dx = e^x + C )
Slide 29:
- Application of Indefinite Integrals
- Indefinite integrals are used to find the most general antiderivative of a function.
- The constant of integration, denoted as ( C ), allows for an infinite family of functions to be represented.
- Indefinite integrals are useful in various fields such as physics, economics, and engineering.
- They allow for the calculation of cumulative values, areas under curves, and the solution of differential equations.
Slide 30:
- Conclusion
- In this lecture, we explored examples of different forms of partial fractions and how to integrate them.
- We also reviewed various integration rules, such as the power rule for integration and the integration of trigonometric and exponential functions.
- Understanding partial fractions and indefinite integrals is an essential skill for success in higher-level mathematics and related disciplines.
- Continue practicing different examples and exploring real-world applications to strengthen your understanding and problem-solving abilities.