Last modified on 27 February 2012, at 17:02

Calculus/Integration techniques/Partial Fraction Decomposition

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Integration techniques/Partial Fraction Decomposition

Suppose we want to find \int {3x+ 1 \over x^2+x} dx. One way to do this is to simplify the integrand by finding constants A and B so that

{3x+ 1 \over x^2+x}={3x+ 1 \over x(x+1)}= {A \over x}+ {B \over x+1}.

This can be done by cross multiplying the fraction which gives

{3x+1\over x(x+1)} = {{A(x+1) + Bx} \over {x(x+1)}}

As both sides have the same denominator we must have

3x+1 = A(x+1)+Bx

This is an equation for x so it must hold whatever value x is. If we put in x=0 we get  1 = A and putting x=-1 gives -2=-B so B=2. So we see that

 \frac{3x+ 1}{x^2+x} = \frac{1}{x} + \frac{2}{x+1} \

Returning to the original integral

 \int \frac{3x+1}{x^2+x} dx =  \int \frac{dx}{x} + \int \frac{2}{x+1} dx
=  \ln \left| x \right| + 2 \ln \left| x+1 \right| + C

Rewriting the integrand as a sum of simpler fractions has allowed us to reduce the initial integral to a sum of simpler integrals. In fact this method works to integrate any rational function.

Method of Partial FractionsEdit

To decompose the rational function \frac{P(x)}{Q(x)}:

1.1).

  • Step 2 Factor Q(x) as far as possible.
  • Step 3 Write down the correct form for the partial fraction decomposition (see below) and solve for the constants.

To factor Q(x) we have to write it as a product of linear factors (of the form ax+b) and irreducible quadratic factors (of the form ax^2+bx+c with b^2-4ac<0).

Some of the factors could be repeated. For instance if Q(x) = x^3-6x^2+9x we factor Q(x) as

 Q(x) = x(x^2-6x+9) = x(x-3)(x-3)=x(x-3)^2.

It is important that in each quadratic factor we have b^2-4ac<0, otherwise it is possible to factor that quadratic piece further. For example if Q(x) = x^3-3x^2 - 2x then we can write

 Q(x) = x(x^2-3x+2) = x(x-1)(x+2)


We will now show how to write P(x)/Q(x) as a sum of terms of the form

 {A \over (ax+b)^k} and {Ax+B \over (ax^2+bx+c)^k}.

Exactly how to do this depends on the factorization of Q(x) and we now give four cases that can occur.

Q(x) is a product of linear factors with no repeatsEdit

This means that Q(x) = (a_1x+b_1)(a_2x+b_2)...(a_nx+b_n) where no factor is repeated and no factor is a multiple of another.

For each linear term we write down something of the form {A \over (ax+b)}, so in total we write

 {P(x) \over Q(x)} = {A_1 \over (a_1x+b_1)} + {A_2 \over (a_2x+b_2)} + \cdots +  {A_n \over (a_nx+b_n)}
Example 1

Find  \int {1+x^2 \over (x+3)(x+5)(x+7)}dx

Here we have P(x)=1+x^2,  Q(x)=(x+3)(x+5)(x+7) and Q(x) is a product of linear factors. So we write

 \frac{1+x^2}{(x+3)(x+5)(x+7)}=\frac{A}{x+3}+\frac{B}{x+5}+\frac{C}{x+7}

Multiply both sides by the denominator

1+x^2=A(x+5)(x+7)+B(x+3)(x+7)+C(x+3)(x+5)

Substitute in three values of x to get three equations for the unknown constants,

\begin{matrix} x=-3 & 1+3^2=2\cdot 4 A \\  x=-5 & 1+5^2=-2\cdot 2 B \\  
x=-7 & 1+7^2=(-4)\cdot (-2) C \end{matrix}

so A=5/4, B=-13/2, C=25/4, and

 \frac{1+x^2}{(x+3)(x+5)(x+7)}=\frac{5}{4x+12} -\frac{13}{2x+10} +\frac{25}{4x+28}

We can now integrate the left hand side.

\int \frac{1+x^2 \, dx}{(x+3)(x+5)(x+7)}=
\frac{5}{4} \ln|x+3| - \frac{13}{2}\ln|x+5|+ \frac{25}{4}\ln|x+7|+ C

ExercisesEdit

Evaluate the following by the method partial fraction decomposition.

1. \int\frac{2x+11}{(x+6)(x+5)}dx

\ln|x+6|+\ln|x+5|+C

2. \int\frac{7x^{2}-5x+6}{(x-1)(x-3)(x-7)}dx

\frac{2}{3}\ln|x-1|-\frac{27}{4}\ln|x-3|+\frac{157}{12}\ln|x-7|+C

Solutions

Q(x) is a product of linear factors some of which are repeatedEdit

If (ax+b) appears in the factorisation of Q(x) k-times then instead of writing the piece  {A \over (ax+b)} we use the more complicated expression

 {A_1\over ax+b} + {A_2\over (ax+b)^2} +  {A_3\over (ax+b)^3} + \cdots + {A_k\over (ax+b)^k}

Example 2

Find  \int {1 \over (x+1)(x+2)^2}dx

Here P(x)=1 and Q(x)=(x+1)(x+2)^2 We write

 \frac{1}{(x+1)(x+2)^2}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}

Multiply both sides by the denominator  1= A(x+2)^2+B(x+1)(x+2)+C(x+1)

Substitute in three values of x to get 3 equations for the unknown constants,

\begin{matrix} x=0 & 1= 2^2A +2B+C \\  x=-1 & 1=A \\  
x=-2 & 1= -C \end{matrix}

so A=1, B=-1, C=-1, and

 \frac{1}{(x+1)(x+2)^2}=\frac{1}{x+1}-\frac{1}{x+2}-\frac{1}{(x+2)^2}

We can now integrate the left hand side.
\int \frac{1}{(x+1)(x+2)^2} dx= \ln \frac{1}{x+1} - \ln \frac{1}{x+2} + \frac{1}{x+2} +C
We now simplify the fuction with the property of Logarithms.
\int \ln \frac{1}{x+1} - \ln \frac{1}{x+2} + \frac{1}{x+2} + C = \ln \frac{x+1}{x+2} + \frac{1}{x+2} +C

ExerciseEdit

3. Evaluate \int\frac{x^{2}-x+2}{x(x+2)^{2}}dx using the method of partial fractions.

\frac{1}{2}\ln|x|+\frac{1}{2}\ln|x+2|+\frac{4}{x+2}+C

Solution

Q(x) contains some quadratic pieces which are not repeatedEdit

If (ax^2+bx+c) appears we use  {Ax+B \over (ax^2+bx+c)}.

ExercisesEdit

Evaluate the following using the method of partial fractions.

4. \int \frac{2}{(x+2)(x^{2}+3)} dx

\frac{2}{7}\ln|x+2|-\frac{1}{7}\ln|x^{2}+3|+\frac{4}{7\sqrt{3}}\arctan(\frac{x}{\sqrt{3}})+C

5. \int\frac{dx}{(x+2)(x^{2}+2)}

\frac{1}{6}\ln|x+2|-\frac{1}{12}\ln|x^{2}+2|+\frac{\sqrt{2}}{6}\arctan(\frac{x}{\sqrt{2}})+C

Solutions

Q(x) contains some repeated quadratic factorsEdit

If (ax^2+bx+c) appears k-times then use

 {A_1x+B_1 \over (ax^2+bx+c)} +  {A_2x+B_2 \over (ax^2+bx+c)^2} + {A_3x+B_3 \over (ax^2+bx+c)^3} + \cdots + {A_kx+B_k \over (ax^2+bx+c)^k}

ExerciseEdit

Evaluate the following using the method of partial fractions.

6. \int\frac{dx}{(x^{2}+1)^{2}(x-1)}

-\frac{1}{2}\arctan(x)+\frac{1-x}{4(x^{2}+1)}+\frac{1}{8}\ln\left(\frac{(x-1)^{2}}{x^{2}+1}\right)+C

Solution

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Integration techniques/Partial Fraction Decomposition