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Integration using Partial Fractions Simplified Revision Notes

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8.2.9 Integration using Partial Fractions

Integration using partial fractions involves breaking down a rational function (a fraction where both the numerator and denominator are polynomials) into simpler fractions that can be integrated individually.

Step-by-step method:

infoNote
  1. Factor the denominator into linear or quadratic terms.
  2. Express the function as a sum of partial fractions.
  3. Solve for constants in the partial fractions.
  4. Integrate each term separately using basic integration rules. This method simplifies complex fractions, making them easier to integrate.
infoNote

Example:

xx216dx\int \frac{x}{x^2 - 16} \, dx

In this question, the idea is to notice the denominator can be factorised as a quadratic, then partial fractions can be used.

1(x+4)(x4)dx\int \frac{1}{(x+4)(x-4)} \, dx

Now split into partial fractions:

1(x+4)(x4)Ax+4+Bx4\frac{1}{(x+4)(x-4)} \equiv \frac{A}{x+4} + \frac{B}{x-4}1A(x4)+B(x+4)\Rightarrow 1 \equiv A(x-4) + B(x+4)

Substitute values to find A and B:

  1. Let x=4x = 4 :
x=41=8BB=18x = 4 \Rightarrow 1 = 8B \Rightarrow B = \frac{1}{8}
  1. Let x=4x = -4 :
x=41=8AA=18x = -4 \Rightarrow 1 = -8A \Rightarrow A = \frac{1}{8}

Therefore:

I=18(x+4)+18(x4)dx181x+4dx+181x4dxI = \int \frac{-1}{8(x+4)}+\frac{1}{8(x-4)} \, dx \equiv \frac{-1}{8} \int \frac{1}{x+4} \, dx + \frac{1}{8} \int \frac{1}{x-4} \, dx=18lnx+4+18lnx4+c= -\frac{1}{8} \ln|x+4| + \frac{1}{8} \ln|x-4| + c=18[lnx4lnx+4]+c=18lnx4x+4+c= \frac{1}{8} [ \ln|x-4| - \ln|x+4| ] + c = \frac{1}{8} \ln \left| \frac{x-4}{x+4} \right| + c

Further Partial Fractions and Integration

Fact: A denominator of x2+ax^2 + a leads to a partial fraction component of Ax+Bx2+a\dfrac{Ax + B}{x^2 + a}.

infoNote

Example: Write 20(x+3)(x2+1)\dfrac{20}{(x+3)(x^2+1)} in partial fractions.

20(x+3)(x2+1)Ax+3+Bx+Cx2+1\frac{20}{(x+3)(x^2+1)} \equiv \frac{A}{x+3} + \frac{Bx+C}{x^2+1}
  1. Multiply through by (x+3)(x2+1)(x+3)(x^2+1):
20=A(x2+1)+(Bx+C)(x+3)20 = A(x^2+1) + (Bx+C)(x+3)
  1. Find constants A, B, and C:
  • Let x=3x = -3:
20=10AA=220 = 10A \Rightarrow A = 2
  • Let x=0x = 0:
20=2(1)+3C18=3CC=620 = 2(1) + 3C \Rightarrow 18 = 3C \Rightarrow C = 6
  • Let x=1x = 1:
20=2(2)+(B(1)+6)(4)20 = 2(2) + (B(1) + 6)(4)20=4+4B+248=4BB=220 = 4 + 4B + 24 \Rightarrow -8 = 4B \Rightarrow B = -220(x+3)(x2+1)2x+3+2x+6x2+1\therefore \frac{20}{(x+3)(x^2+1)} \equiv \frac{2}{x+3} + \frac{-2x+6}{x^2+1}

infoNote

Example: By expressing x4+xx4+5x2+6\dfrac{x^4 + x }{x^4 + 5x^2 + 6} in partial fractions, find

x4+xx4+5x2+6dx\int \frac{x^4 + x }{x^4 + 5x^2 + 6} \, dxx4+5x2+65x2+x6x4+5x2+6 dx\int \frac {x^4 + 5x^2 + 6-5x^2 + x - 6}{x^4 + 5x^2 + 6}\ dx=(15x2x+6x4+5x2+6)dx= \int \left(1 - \frac{5x^2 - x + 6}{x^4 + 5x^2 + 6}\right) \, dx5x2x+6(x2+3)(x2+2)=Ax+Bx2+3+Cx+Dx2+2\frac{5x^2 - x + 6}{(x^2+3)(x^2+2)} = \frac{Ax+B}{x^2+3} + \frac{Cx+D}{x^2+2}
  1. Multiply both sides by (x2+3)(x2+2)(x^2+3)(x^2+2):
5x2x+6=(Ax+B)(x2+2)+(Cx+D)(x2+3)5x^2 - x + 6 = (Ax+B)(x^2+2) + (Cx+D)(x^2+3)
  1. Expand both sides:
5x2x+6=Ax3+Bx2+2Ax+2B+Cx3+Dx2+3Cx+3D5x^2 - x + 6 = A x^3 + Bx^2 + 2Ax + 2B + Cx^3 + Dx^2 + 3Cx + 3D
  1. Combine like terms:
5x2x+6=x3(A+C)+x2(B+D)+x(2A+3C)+(2B+3D)5x^2 - x + 6 = x^3(A+C) + x^2(B+D) + x(2A+3C) + (2B+3D)
  1. Set up equations by equating coefficients:
  • A+C=0A + C = 0
  • B+D=5B + D = 5
  • 2A+3C=12A + 3C = -1
  • 2B+3D=62B + 3D = 6

  1. Solve the equations:
  • From A+C=0,C=AA + C = 0, C = -A.
  • From 2A+3C=12A + 3C = -1, substituting C=AC = -A:
2A+3(A)=1A=1A=1,C=12A + 3(-A) = -1 \Rightarrow -A = -1 \Rightarrow A = 1, C = -1
  • From B+D=5B + D = 5:
2B+3D=62B+3(5B)=62B + 3D = 6 \Rightarrow 2B + 3(5-B) = 62B+153B=6B+15=6B=9,D=42B + 15 - 3B = 6 \Rightarrow -B + 15 = 6 \Rightarrow B = 9, D = -45x2x+6(x2+3)(x2+2)=x+9x2+3+x+4x2+2\therefore \frac{5x^2 - x + 6}{(x^2+3)(x^2+2)} = \frac{x+9}{x^2+3} + \frac{-x+4}{x^2+2}
  1. Integrate:
5x2x+6(x2+3)(x2+2)dx=x+9x2+3dx+x+4x2+2dx\int \frac{5x^2 - x + 6}{(x^2+3)(x^2+2)} dx = \int \frac{x+9}{x^2+3} dx + \int \frac{-x+4}{x^2+2} dxxx2+3dx + 9x2+3dx  xx2+2dx  4x2+2dx\int \frac{x}{x^2+3} dx \ + \ \int \frac{9}{x^2+3} dx \ - \ \int \frac{x}{x^2+2} dx \ - \ \int \frac{4}{x^2+2} dx 122xx2+3dx + 9x2+3dx  122xx2+2dx  124x2+2dx\frac {1}{2} \int \frac {2x}{x^2+3} dx \ + \ \int \frac {9}{x^2+3} dx \ - \ \frac {1}{2} \int \frac {2x}{x^2+2} dx \ - \ \frac {1}{2} \int \frac {4}{x^2+2} dx ABCD122xx2+3dx9x2+3dx122xx2+2dx124x2+2dx\def\arraystretch{1.5} \begin{array}{c:c:c:c} \hline A & B & C & D \\ \hline \frac {1}{2} \int \frac {2x}{x^2+3} dx & \int \frac {9}{x^2+3} dx & \frac {1}{2} \int \frac {2x}{x^2+2} dx & \frac {1}{2} \int \frac {4}{x^2+2} dx \\ \hline \end{array}
  • Let I=x+9x2+3dx+x+4x2+2dxI = \int \dfrac{x+9}{x^2+3} dx + \int \dfrac{-x+4}{x^2+2} dx
  • A=12lnx2+3+c1A = \frac{1}{2}\ln|x^2+3| + c_1
  • B=93arctan(x3)+c3B = \frac{9}{\sqrt{3}} \arctan\left(\frac{x}{\sqrt{3}}\right) + c_3
  • C=12lnx2+2+c2C = \frac{1}{2} \ln|x^2+2| + c_2
  • D=42arctan(x2)+c4D = \frac{4}{\sqrt{2}} \arctan\left(\frac{x}{\sqrt{2}}\right) + c_4
I=x[12lnx2+3+93arctan(x3)12lnx2+2+42arctan(x2)]+c\therefore \quad I =\\ x- \left [\frac{1}{2}\ln|x^2+3| + \frac{9}{\sqrt{3}}\arctan\left(\frac{x}{\sqrt{3}}\right) - \frac{1}{2}\ln|x^2+2| + \frac{4}{\sqrt{2}}\arctan\left(\frac{x}{\sqrt{2}}\right) \right ] + c=x[12ln(x2+3x2+2)+33arctan(3x3)22arctan(2x2)]+c=x - \left [ \frac{1}{2}\ln\left(\frac{x^2+3}{x^2+2}\right) + 3\sqrt{3}\arctan\left(\frac{\sqrt{3}x}{3}\right) - 2\sqrt{2}\arctan\left(\frac{\sqrt{2}x}{2}\right) \right ]+ c

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