Partial Fractions In Integration - Calculus

Any number which can be easily represented in the form of p/q, such that p and q are integers and q≠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)≠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 improper rational function. With the help of 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:

\(\frac{P(x)}{Q(x)}= A(x) + \frac{R(x)}{Q(x)}\)

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:

∫f(x)dx = F(x) + C

Here R.H.S. of the equation means integral of f(x) with respect to x.

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. The following table explains the conversion of basic rational functions into partial fractions.


Rational Function

Partial Function


\(\large \frac{p(x) + q}{(x-a)(x-b)}\)

\(\large \frac{A}{x-a} + \frac{B}{(x-b)}\)


\(\large \frac{p(x) + q}{(x-a)^{2}}\)

\(\large \frac{A}{x-a} + \frac{B}{(x-b)^{2}}\)


\(\large \frac{px^{2} + qx +r}{(x-a)(x-b)(x-c)}\)

\(\large \frac{A}{x-a} + \frac{B}{(x-b)} + \frac{B}{(x-c)}\)


\(\large \frac{px^{2} + q(x) +r}{(x-a)^{2}(x-b)} \)

\(\large \frac{A}{x-a} + \frac{B}{(x-a)^{2}} + \frac{B}{(x-b)}\)


\(\large \frac{px^{2} + qx +r}{(x-a)(x^{2}+bx+c)}\)

\(\large \frac{A}{x-a} + \frac{Bx+c}{x^{2}+bx+c}\)

Here A, B and C are real numbers.

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)


Equating the coefficients of x and the constant, we have

A + B = 0

A – 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.,

∫[f(x) +g(x)]dx = ∫f(x)dx + ∫g(x)dx


\(= \frac{1}{4} \int \frac{1}{(x-3)} – \frac{1}{4} \int \frac{1}{(x+1)}\)

\(= \frac{1}{4} \ln \left | x-3 \right | – \frac{1}{4} \ln \left | x+1 \right |\)

\(= \frac{1}{4} \ln \left | \frac{x-3}{x+1} \right |\)<

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x + 5(x  2)2dx