Partial Fraction Expansion
If f(x) is a function that is required to be integrated, f(x) is called the Integrand, and the integration of the function without any limits or boundaries is known as the Indefinite Integration. Indefinite integration has its own formulae to make the process of integration easier. However, sometimes there are functions that are too complex and can not be easily integrated and require more time. For example, having a long quadratic expression in the denominator of the function. In such cases, indefinite integration can be done with the help of partial fractions in order to make the process easy and less time-consuming.
Indefinite Integration
Indefinite Integrations are also known as Anti-derivatives, as it is the reverse process of differentiation. Indefinite means that the function is integrated without any limits or boundaries and the entire function is required to be integrated. The function, say f(x), is the Integrand, x here is the variable of the integration, and the symbol used for integration is âˆ«. Let’s take a look at some basic formulae used for indefinite integration,
Some basic formulae of indefinite integration:
Functions | Formulae |
âˆ«a dx | ax + C, aâˆˆ R |
âˆ«x^{n} dx | +C, xâˆ‰ -1 |
âˆ«1/x dx | log|x| + C |
âˆ«e^{x} dx | e^{x} + C |
âˆ«1/âˆšx dx | 2âˆšx + C |
âˆ«a^{x} dx | |
âˆ«Sinx dx | -Cosx +C |
âˆ«Cosx dx | Sinx +C |
âˆ«Tanx dx | log|secx| +C |
âˆ«Secx tanx dx | Secx + C |
âˆ«Cosecx cotx dx | -Cosecx + C |
âˆ« Sec^{2}x dx | tanx +C |
âˆ« Cosec^{2}x dx | – cotx + C |
Example 1: Find the integration of the function,
f(x) = x^{5} + 3/x
Solution:
âˆ«f(x)dx = âˆ«x^{5 }dx + âˆ«3/x dx
âˆ«f(x)dx =
Example 2: Integrate, f(x)= 5cosx – 9tanx
Solution:
âˆ«f(x)dx= âˆ«[5cosx- 9tanx]dx
âˆ«f(x)dx= âˆ«5cosxdx- âˆ«9tanxdx
âˆ«f(x)dx= 5sinx- (log|secx|) +C
Apart from the simple problems that only require the formulae, it is important to understand that complex functions can not be solved this easily. In order to find the integration of the complex function, one method is to use Partial Fractions,
Integration by Partial Fraction
The rational functions are defined as the functions that are in the form of P(x)/Q(x) where Q(x) â‰ 0. These functions can be either in proper form or Improper form. The Proper ration function is defined as the one where the highest degree in Q(x) is more than in P(x). Improper functions are those in which P(x) has its highest degree more than Q(x). The rational functions can be integrated by Partial fraction very easily provided that the denominator in the function can be factorized into linear factors.
Partial Fraction Decomposition
The integrand, that is, the given function can be factorized into simpler form and this is called as partial fraction decomposition. A few generalized Partial fractions of the functions is given below,
Functions | Partial Fractions |
, aâ‰ b | |
, x^{2}+ bx+ c cannot be factorized |
How are these generalized forms obtained or are there any steps that can be used for partial fraction decomposition? YES. The steps can easily help in reaching a perfect partial fraction form,
Steps for Partial Fraction Decomposition
- In order to decompose, start with the proper rational expression. Factorize the denominator into the most basic form.
- Write down the separate partial fractions obtained and, in order to deal with the numerator, fill them with variables that will be found out soon.
- Now, in order to find the value of the variables, here, A, B, and C, multiply the equation by the denominator.
- Solve the variables by substituting different values accordingly.
- When the values of the numerators are obtained, place them in the partial fraction.
Integration using Long Division or Synthetic Division
This method is used when the functions are not given in the rational form, that is, the fractions are improper fractions and the degree of the numerator is more or equal than the degree of the denominator. In these cases, suppose the degree in the numerator is a and the degree in the denominator is b, while doing partial fractions here, it needs to be kept in mind that an additional value of (a-b) is added to the denominators along with the partial fractions.
Let us consider the function given below:
f(x) =
This function is an improper integral, thus it can be solved using the long division method.
Here, S(x) = x^{5} – 6x^{4} + 5x^{2} + 8 and R(x) = -100 and Q(x) = x + 2.
So, the integral can be re-written as,
âˆ«(x^{5} – 6x^{4} + 5x^{2} + 8 + )dx
â‡’
If the numerator P(x) has a degree greater than or equal to the degree of the denominator Q(x), then the rational function is called improper. In this case, we use long division of polynomials to write the ratio as a polynomial with a remainder.
Let’s say dividing P(x) by Q(x) gives S(x) with the remainder R(x), then the degree of R(x) is less than the degree of Q(x) as a result of the long division.
After doing this, the function can be integrated.
Sample Problems
Question 1: Integrate, f(x)= sin(x) + 3tanx
Solution :
âˆ«f(x)dx= âˆ«[sin(x) + 3tanx]dx
âˆ«f(x)dx= âˆ«sin(x)dx + âˆ«3tanxdx
âˆ«f(x)dx= -cos(x) + 3(log|secx|) +C
Question 2: Find the integral of the following function.
f(x) =
Solution:
The given function can be decomposed into partial fractions.
f(x) =
Comparing both sides of the equation,
A + B = 1
2A = 1
From both these equations, it can be concluded that.
A = 1/2, B = 1/2
Thus, the function becomes,
f(x) =
Now,
F(x) = âˆ«f(x)
â‡’ F(x) =
â‡’ F(x) =
Question 3: Find the integral of the following function.
f(x) =
Solution:
The given function can be decomposed into partial fractions.
f(x) =
Comparing both sides of the equation,
A + B = 1
2A + B = 0
From both these equations, it can be concluded that.
A = -1, B = 2
Thus, the function becomes,
f(x) =
Now,
F(x) = âˆ«f(x)
â‡’ F(x) =
â‡’ F(x) =
Question 4: Find the integral of the following function.
f(x) =
Solution:
The given function can be decomposed into partial fractions. First let’s simplify the function.
f(x) =
Decomposing the function,
Comparing both sides of the equation,
A + B = 0
A = 1
From both these equations, it can be concluded that.
A = 1, B = -1
Thus, the function becomes,
f(x) =
Now,
F(x) = âˆ«f(x)
â‡’ F(x) =
â‡’ F(x) = ln(x) – ln(x+1)
Question 5: Find the integral of the following function.
f(x) =
Solution:
The given function can be decomposed into partial fractions. First let’s simplify the function.
f(x) =
Integrating the function,
F(x) = âˆ«f(x)dx
â‡’ F(x) = âˆ«()dx
â‡’ F(x) =
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