Partial fractions and integration

Multiply both sides of $\frac{1}{(x-4)(x+4)}=\frac{A}{x-4}+\frac{B}{x+4}$ by $(x-4)(x+4)$ to get $1=A(x+4)+B(x-4)$. Setting $x=4$ gives $A=\frac{1}{8}$, and setting $x=-4$ gives $B=-\frac{1}{8}$.

After splitting, integrate term by term.

This uses $\int\frac{1}{x-a}\,dx=\log|x-a|+C$, so keep the absolute value bars.

Examples

1. 01Integrate:$\displaystyle \int \left(\frac{1}{x-3} - \frac{1}{x+2}\right)dx$AnswerHide

   Answer$\displaystyle \log_e|x-3| - \log_e|x+2|+C$

   Steps

   1. Both pieces have the shape $\frac{1}{x}$, so both become $\log$, and the subtraction stays a subtraction
   2. $\displaystyle \int \left(\frac{1}{x-3}-\frac{1}{x+2}\right)dx$$\displaystyle = \log_e|x-3|-\log_e|x+2|+C$
   3. Do not turn the difference into a sum; that is a different function
2. 02What is A?$\displaystyle \frac{1}{(x-4)(x+4)} = \frac{A}{x-4} + \frac{B}{x+4}$AnswerHide

   Answer$\displaystyle \frac{1}{8}$

   Steps

   1. Multiply both sides by $x-4$: the left becomes $\frac{1}{x+4}$. Now put $x=4$ in
   2. $\displaystyle \frac{1}{4+4}$$\displaystyle = \frac{1}{8}$
   3. $B$ works the same way, but with $x=-4$, so its sign is opposite
3. 03What goes in □?$\displaystyle \frac{2}{(x-4)(x+1)} = \frac{A}{x-4} + \frac{B}{\square}$AnswerHide

   Answer$\displaystyle x+1$

   Steps

   1. The denominators you split into are exactly the factors, taken one at a time
   2. The factors of $(x-4)(x+1)$ are $x-4$ and $x+1$; signs and powers stay as they are
4. 04Integrate:$\displaystyle \int \left(\frac{1}{x-2} - \frac{1}{x+4}\right)dx$AnswerHide

   Answer$\displaystyle \log_e|x-2| - \log_e|x+4|+C$

   Steps

   1. Both pieces have the shape $\frac{1}{x}$, so both become $\log$, and the subtraction stays a subtraction
   2. $\displaystyle \int \left(\frac{1}{x-2}-\frac{1}{x+4}\right)dx$$\displaystyle = \log_e|x-2|-\log_e|x+4|+C$
   3. Do not turn the difference into a sum; that is a different function
5. 05Integrate:$\displaystyle \int \left(\frac{1}{x-4} - \frac{1}{x+2}\right)dx$AnswerHide

   Answer$\displaystyle \log_e|x-4| - \log_e|x+2|+C$

   Steps

   1. Both pieces have the shape $\frac{1}{x}$, so both become $\log$, and the subtraction stays a subtraction
   2. $\displaystyle \int \left(\frac{1}{x-4}-\frac{1}{x+2}\right)dx$$\displaystyle = \log_e|x-4|-\log_e|x+2|+C$
   3. Do not turn the difference into a sum; that is a different function
6. 06What is A?$\displaystyle \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$AnswerHide

   Answer$\displaystyle \frac{1}{2}$

   Steps

   1. Multiply both sides by $x-1$: the left becomes $\frac{1}{x+1}$. Now put $x=1$ in
   2. $\displaystyle \frac{1}{1+1}$$\displaystyle = \frac{1}{2}$
   3. $B$ works the same way, but with $x=-1$, so its sign is opposite

FAQ

Q1What is a quick way to find $A$ and $B$?AnswerHide

A

In the equation cleared of denominators, substitute the $x$ that makes a factor zero. In $1=A(x+4)+B(x-4)$, setting $x=4$ removes the $B$ term and leaves only $A$. Comparing coefficients gives the same answer.

Q2What if the numerator's degree is at least the denominator's?AnswerHide

A

Divide first, to get a polynomial plus a fraction whose numerator has lower degree than the denominator. Split into partial fractions after that.