Calculus/Integration techniques/Partial Fraction Decomposition

< Calculus‎ | Integration techniques
← Integration techniques/Trigonometric Integrals Calculus Integration techniques/Tangent Half Angle →
Integration techniques/Partial Fraction Decomposition

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

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

This can be done by cross multiplying the fraction which gives

\frac{3x+1}{x(x+1)}=\frac{A(x+1)+Bx}{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 A=1 and putting x=-1 gives =-B=-2 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
=\int\frac{dx}{x}+2\int\frac{dx}{x+1}
=\ln|x|+2\ln\Big|x+1\Big|+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.

Contents

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 \frac{P(x)}{Q(x)} as a sum of terms of the form

\frac{A}{(ax+b)^k} and \frac{Ax+B}{(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)\cdots(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 \frac{A}{(ax+b)} , so in total we write

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

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

ExercisesEdit

Evaluate the following by the method partial fraction decomposition.

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

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

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

\tfrac{2}{3}\ln\Big|x-1\Big|-\tfrac{27}{4}\ln\Big|x-3\Big|+\tfrac{157}{12}\ln\Big|x-7\Big|+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 \frac{A}{ax+b} we use the more complicated expression

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

Example 2

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

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{dx}{(x+1)(x+2)^2}=\ln\left|\frac{1}{x+1}\right|-\ln\left|\frac{1}{x+2}\right|+\frac{1}{x+2}+C

We now simplify the fuction with the property of Logarithms.

\ln\left|\frac{1}{x+1}\right|-\ln\left|\frac{1}{x+2}\right|+\frac{1}{x+2}+C=\ln\left|\frac{x+2}{x+1}\right|+\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{\ln\Big|x(x+2)\Big|}{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 \frac{Ax+B}{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{\ln\left(\frac{(x+2)^2}{x^2+3}\right)}{7}+\frac{4\arctan\Big(\tfrac{x}{\sqrt3}\Big)}{7\sqrt3}+C

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

\frac{\ln\left(\frac{(x+2)^2}{x^2+2}\right)+2\sqrt2\arctan\Big(\tfrac{x}{\sqrt2}\Big)}{12}+C

Solutions

Q(x) contains some repeated quadratic factorsEdit

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

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

ExerciseEdit

Evaluate the following using the method of partial fractions.

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

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

Solution

← Integration techniques/Trigonometric Integrals Calculus Integration techniques/Tangent Half Angle →
Integration techniques/Partial Fraction Decomposition