Subject: Calculus

# Partial Fractions In Integration

Another topic in integration that I would like to share with you in this article is about partial fractions in integration. This technique is employed when the integrand, usually in the form of an algebraic fraction, can never be integrated using direct methods. Examples of such integrals can be classified into two types: reducible and irreducible types of polynomial in the denominator.

## Reducible Polynomial in the Denominator

Reducible Polynomial, as its name suggests, is when the denominator in the fraction (integrand) can be reduced into a form that is easy to integrate. For example, the form

\int \frac{dx}{\left (ax+b\right )^n}
,

if we let u=ax+b such that du=adx, the integral above would then be reduced into

\int \frac{du}{a\left (u \right )^n}

which can now easily be integrated using direct methods. Proceeding with integration, we see that

\int \frac{du}{a\left (u \right )^n} = \frac{1}{a} \int u^{-n}du = \frac{1}{a} \cdot \frac{u^{-n+1}}{-n+1} + C

Therefore,

\int \frac{dx}{\left (ax+b\right )^n} = \frac{1}{a}\cdot \frac{\left (ax+b \right )^{1-n}}{1-n} + C

if and only if n \neq 1. But if n= 1, then

\int \frac{dx}{\left (ax+b\right )^n} = \frac{1}{a} \int \frac{du}{u}
=\frac{1}{a} \ln{u} + C

Thus,

\int \frac{dx}{\left (ax+b\right )^n} = \frac{1}{a} \ln{\left |ax+b \right |} + C

## Irreducible Polynomial in the Denominator

Integrals with this type of integrands are more difficult than the ones above due to the fact that it cannot be reduced to a form directly like what we did above. What we do here is we express the fraction into a simplier fraction by writing any rational function of a real variable into a sum of a polynomial function and a finite number of fractions. For example, consider a sum of algebraic fractions

\frac{3}{3x+1} + \frac{2}{2x+5} = \frac{3\left (2x+5 \right ) + 2\left (3x+1 \right )}{\left (2x+5\right ) \left (3x+1 \right )} = \frac{12x+17}{6x^2+17x+5}

so if we have an integrand of the form \frac{12x+17}{6x^2+17x+5}, all we have to do is to transform it to a form \frac{3}{3x+1} + \frac{2}{2x+5} which is now easy to integrate. This transformation can be done by doing the reverse process of what we did above. So basically, that's what integration in partial fractions is all about.

Actually, there are three forms of integrals with irreducible polynomials in the denominator. These are:

### Denominators with Linear Factors

Integrals with denominators with linear factors takes the form below when f(x) is any function in x with degree lower than that of the denominator,

\int \frac{f(x)}{\left ( x+a \right ) \left ( x+b \right )}dx
.

To "break" this fraction into a sum of polynomial function that can be integrated easily, we employ a transformation technique which can be done by writing the integrand as the one below;

\frac{f(x)}{\left ( x+a \right ) \left ( x+b \right )} = \frac{A}{x+a}+\frac{B}{x+b}

where A and B are functions that are yet to be determined.

#### Example

Transform the following fraction into a sum of polynomial function using the technique above.

\frac{3x+11}{\left ( x-3 \right ) \left ( x+2 \right )}

The fraction above can be written as

\frac{3x+11}{\left ( x-3 \right ) \left ( x+2 \right )} = \frac{A}{x-3}+\frac{B}{x+2}
= \frac{A\left ( x+2\right ) + B\left (x-3 \right )}{\left ( x-3 \right ) \left ( x+2 \right )}= \frac{\left (A+B\right )x + 2A-3B}{\left ( x-3 \right ) \left ( x+2 \right )}

Then, we can deduce that

3x+11 = \left (A+B\right )x + 2A-3B

from then where we can get

A= 4

and

B=-1

Thus, the integrand has successfully been transformed into

\frac{3x+11}{\left ( x-3 \right ) \left ( x+2 \right )} = \frac{4}{x-3}-\frac{1}{x+2}

which can now easily be integrated directly.

When the term in the denominator is quadratic and is factorable, we have the case above. But when it is not factorable, then we have the following form such that f(x) is of degree 1,

\int \frac{f(x)}{ax^2+bx+c}dx

To transform this into sum of simpler fractions, all we have to do is to complete the squares in the denominator and proceed accordingly or by following the formulation below and proceed with the same steps above.

\frac{f(x)}{ax^2+bx+c} = \frac{Ax+B}{ax^2+bx+C}

## In General...

As a summary, I will give the possible forms of the integrand and its corresponding partial fractions solution.

### When the denominator has the following form...

1. ax+b
2. (ax+b)^n
3. ax^2+bx+c
4. (ax^2+bx+c)^n

### then its corresponding partial fraction decomposition is

1. \frac{A}{ax+b}
2. \frac{A_1}{ax+b}+\frac{A_2}{\left (ax+b\right )^2}+\frac{A_3}{\left (ax+b\right )^3}+\cdot \cdot \cdot +\frac{A_n}{\left (ax+b\right )^n}
where n = 1, 2, 3, ...
1. \frac{Ax+B}{ax^2+bx+c}
2. \frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{\left (ax^2+bx+c\right )^2} + \frac{A_3x+B_3}{\left (ax^2+bx+c\right )^3}+\cdot \cdot \cdot \frac{A_nx+B_n}{\left (ax^2+bx+c\right )^n}
where n = 1, 2, 3, ...