Partial Fraction Decomposition

Unit 7.4 of Math 152

General form

• Suppose

Examples

• Example 3.11 b:

  • $I=\int \frac{x^3-4x-1}{x(x-1{)}^3}dx$
    • since there is a multiplicity 3 you must stretch it until the sum is the fraction
      • -> $\frac{A}{x}+\frac{B}{(x-1)}+\frac{C}{(x-1{)}^2}+\frac{D}{(x-1{)}^3}$
        • => $x^3-4x-1=(x-1{)}^{3}A+x(x-1{)}^{2}B+x(x-1)C+xD$
          • multiplying through the denominator
          • [x^3] 1=A+B =>B=0
          • [x^2] 0=-3A-2B+C => C=3
          • [x^1] -4 =3A+b-C+d => D=-4
          • [x^0] -1=-A => A=1
            • => the numbers in the front is from beginning in the coefficients of the corresponding multiplicity
              • $I=\int \frac{1}{x}dx+\int \frac{3}{(x-1{)}^2}dx+\int -\frac{4}{(x-1{)}^3}dx$
                • => $\ln |x|-\frac{3}{}$

• Example c.

  • $I=\int \frac{5x^3-3x^2+2x-1}{x^4+x^2}dx$ => we can’t fully factor because its unfactorable quadratic => $x^2(x^2+1)$
    • => $\frac{A}{x}+\frac{B}{x^2}+\frac{C+Dx}{x^2+1}$
      • =>$5x^3-3x^2+2x-1⟹x(x+1)A+(x^2+1)B+x^{2}C+x^{3}D$
      • [x^3] 5= A+D -> D=3
      • -3= B+C -> C=-2
      • 2=A
      • -1 = B
        • $I+2\ln |x|+\frac{1}{x}-2\arctan x+\frac{3}{2}\ln (x^2+1)+C$

• Example d.

  • $\int \frac{1}{x(x^2+1{)}^2}dx$
    • => $\frac{A}{x}+\frac{B+Cx}{x^2+1}+{\frac{D+Ex}{x^2+1}}^2$
      • left to you
        • A=1
        • B=0
        • C=

• Example E

  • => using Volume of integrals Cylinders
    • => $V={\int }_1^{2}2\pi x\left(\frac{x-9}{x^2-3x}\right)dx$
      • =>doing polynomial division
        • -> $\frac{x-9}{x-3}=1+\frac{-6}{x-3}$
          • =>$V={\int }_1^{2}2\pi \left(1-\frac{6}{x-3}\right)dx$