Partial Fractions

Home

Linear Factors

For every factor of $(x-a)$ in $Q(x)$, we have a term $\displaystyle\frac{A}{x-a}$.

For every repeated linear factor $(x-a)^n$, we have $$\displaystyle\frac{A_1}{(x-a)} + \frac{A_2}{(x-a)^2} + \ldots + \frac{A_n}{(x-a)^n}.$$

The video below goes over some examples.

More Examples

Example 1: Find $\displaystyle\int\frac6{x^2-1}\,dx$.

Solution: Since $ x^2-1= (x-1)(x+1)$, we look for a decomposition of the form $$ \displaystyle{ 6 \over (x-1)(x+1) } = \displaystyle{ { A \over x-1 } + { B \over x+1 } } $$ for some $A,\,B$. Taking the common denominator, we have $$ \displaystyle{ 6 \over (x-1)(x+1) } = \displaystyle{ { A(x+1) +B(x-1)\over (x-1)(x+1) }}. $$ This is true if the numerators are equal, i.e., $$ 6=A(x+1)+B(x-1).$$ To find out $A$, substitute $x=1$, then $6=A(2)+B(0)$, so $A=3$. Similarly, substituting $x=-1$ lets us find that $B=-3$. Finally, we have $$ \int \displaystyle{ 6 \over (x-1)(x+1) }\, dx= \int \displaystyle{ {3 \over x-1}\,dx +\int {-3 \over x+1}\,dx }=3\log{(x-1)}-3\log{(x+1)}+ C. $$

Example 2: Find the partial fraction decomposition of $$ \frac{1}{x^4-3x^3+3x^2-x}. $$

Solution: Since $Q(x)=x\,\left(x-1\right)^3$, we have both a non-repeated factor, $x$, and a repeated factor, $(x-1)^3$, so we're looking for a decomposition of the form $$ \frac{1}{x\,\left(x-1\right)^3}=\frac{A}{x}+\frac{B_1}{x-1}+\frac{B_2}{\left(x-1\right)^2}+\frac{B_3}{\left(x-1\right)^3}. $$ Taking the common denominator, we have $$ \frac{1}{x\,\left(x-1\right)^3}=\frac{A\left(x-1\right)^3+B_1x\,(x-1)^2+B_2x\,\left(x-1\right)+B_3x} {x\left(x-1\right)^3}. $$ This is true if the numerators are equal, i.e., $$ 1=A\left(x-1\right)^3+B_1x\,(x-1)^2+B_2x\,\left(x-1\right)+B_3x.$$ To find out $B_3$, substitute $x=1$, then $1=0+B_3$, so $B_3=1$. Similarly, substituting $x=0$ let us find that $A=-1$. To find the other coefficients, we need to choose other values for $x$. For example, taking $x=2$ yields $$ 1=A+2B_1+2B_2+2B_3=-1+2B_1+2B_2+2, $$ i.e., $$ B_1+B_2=0. $$ Taking $x=-1$ yields $$ 1=-8A-4B_1+2B_2-B_3=8-4B_1+2B_2-1, $$ i.e., $$ -2B_1+B_2=-6. $$ Solving the system $$ \begin{cases} B_1+B_2=0\\ -2B_1+B_2=-6 \end{cases} $$ yields $B_1=2$ and $B_2=-2$, so that our decomposition becomes $$ \frac{1}{x\,\left(x-1\right)^3}=-\frac{1}{x}+\frac{2}{x-1}-\frac{2}{\left(x-1\right)^2}+\frac{1}{\left(x-1\right)^3}. $$

The Fundamental Theorem of Calculus

The Indefinite Integral and the Net Change

Substitution

Area Between Curves

Volumes

Integration by Parts

Integrals of Trig Functions

Trig Substitutions

Partial Fractions

Strategies of Integration

Improper Integrals

Differential Equations

Models of Growth

Infinite Sequences

Infinite Series

Integral Test

Comparison Tests

Convergence of Series with Negative Terms

The Ratio and Root Tests

Strategies for testing Series

Power Series

Representing Functions as Power Series

Taylor and Maclaurin Series

Applications of Taylor Polynomials

Partial Derivatives

Multiple Integrals

Iterated Integrals over Rectangles

Double Integrals over General Regions

Linear Factors

More Examples