An algebraic fraction can be broken down into simpler parts known as “partial fractions“. Consider an algebraic fraction, (3x+5)/(2x^{2}-5x-3). This expression can be split into simple form like ((2)/(x-3))-((1)/(2x+1))
The Simpler parts ((2)/(x-3))-((1)/(2x+1))Â are known as partial fractions.
This means that an algebraic expression can be written in the form of
(3x+5)/(2x^{2}-5x-3) =Â ((2)/(x-3))-((1)/(2x+1))
Partial Fractions From Rational Functions
Any number which can be easily represented in the form of p/q, such that p and q are integers and qâ‰ 0 are known as Rational numbers. Similarly, we can define a rational function as the ratio of two polynomial functions P(x) and Q(x), where P and Q are polynomials in x and Q(x)â‰ 0. A rational function is known as proper if the degree of P(x) is less than the degree of Q(x) otherwise it is known as an improper rational function. With the help of the long division process, we can reduce improper rational functions to proper rational functions. Therefore, if P(x)/Q(x) is improper then it can be expressed as:
\(\frac{P(x)}{Q(x)}= A(x) +Â \frac{R(x)}{Q(x)}\)
Here, A(x) is a polynomial in x and R(x)/Q(x) is a proper rational function.
We know that the integration of a function f(x) is given by F(x) and it is represented by:
âˆ«f(x)dx = F(x) + C
Here R.H.S. of the equation means integral of f(x) with respect to x.
Partial Fractions Decomposition
In order to integrate a rational function, it is reduced to a proper rational function. This method in which the integrand is expressed as the sum of simpler rational functions is known as decomposition into partial fractions. After splitting the integrand into partial fractions, it is integrated accordingly with the help of traditional integrating techniques. Here the list of Partial fractions formulas are given
S.No | Rational Function | Partial Function |
1 | \(\large \frac{p(x) + q}{(x-a)(x-b)}\) | \(\large \frac{A}{x-a} + \frac{B}{(x-b)}\) |
2 | \(\large \frac{p(x) + q}{(x-a)^{2}}\) | \(\large \frac{A_{1}}{x-a} + \frac{A_{2}}{(x-a)^{2}}\) |
3 | \(\large \frac{px^{2} + qx +r}{(x-a)(x-b)(x-c)}\) | \(\large \frac{A}{x-a} + \frac{B}{(x-b)} + \frac{C}{(x-c)}\) |
4 | \(\large \frac{px^{2} + q(x) +r}{(x-a)^{2}(x-b)} \) | \(\large \frac{A_{1}}{x-a} + \frac{A_{2}}{(x-a)^{2}} + \frac{B}{(x-b)}\) |
5 | \(\large \frac{px^{2} + qx +r}{(x-a)(x^{2}+bx+c)}\) | \(\large \frac{A}{x-a} + \frac{Bx+C}{x^{2}+bx+c}\) |
Here A, B and C are real numbers.
Partial Fraction of Improper Fraction
An algebraic fraction is improper if the degree of the numerator is greater than or equal to that of the denominator. The degree is the highest power of the polynomial. Suppose, m is the degree of the denominator and n is the degree of the numerator. Then, in addition to the partial fractions arising from factors in the denominator, we must include an additional term: this additional term is a polynomial of degree n âˆ’ m.
Note:Â
- A polynomial with zero degree is K, where K is a constant
- A polynomial of degree 1 is Px + Q
- A polynomial of degree 2 is Px^{2}+Qx+K
Partial Fraction Integration Example
Let us look into an example to have a better insight of integration using partial fractions.
Example: Integrate the function \(\frac{1}{(x-3)(x+1)}\) with respect to x.
Solution: The given integrand can be expressed in the form of partial fraction as:
\(\frac{1}{(x-3)(x+1)} = \frac{A}{(x-3)} + \frac{B}{(x+1)}\)
To determine the value of real coefficients A and B, the above equation is rewritten as:
1= A(x+1)+B(x-3)
â‡’1=x(A+B)+A-3B
Equating the coefficients of x and the constant, we have
A + B = 0
A â€“ 3B = 1
Solving these equations simultaneously, the value of A =1/4 and B = -1/4. Substituting these values in the equation 1, we have
\(\frac{1}{(x-3)(x+1)} = \frac{1}{4(x-3)} + \frac{-1}{4(x+1)}\)
Integrating with respect to x we have;
\(\int \frac{1}{(x-3)(x+1)} = \int \frac{1}{4(x-3)} + \int \frac{-1}{4(x+1)}\)
According to the properties of integration, the integral of sum of two functions is equal to the sum of integrals of the given functions, i.e.,
âˆ«[f(x) +g(x)]dx =Â âˆ«f(x)dx +Â âˆ«g(x)dx
Therefore,
\(= \frac{1}{4} \int \frac{1}{(x-3)} – \frac{1}{4} \int \frac{1}{(x+1)}\)
\(= \frac{1}{4} \ln \left | x-3 \right | – \frac{1}{4} \ln \left | x+1 \right |\)
\(= \frac{1}{4} \ln \left | \frac{x-3}{x+1} \right |\)
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IntegrationÂ | Integration by Substitution |
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