High School Calculus/Integration by Partial Fractions

Integration by Partial Fractions

There are a couple different methods of integrating partial fractions. The one I will be showing you is the one I am most comfortable with.

The important thing is to realize how to split up the denominator into two or more fractions. The numerator will have place holders like A, B, C, and so forth. You may have a Ax + B or some variation thereof depending on how the denominator needs to be split.

Because I think this will be better explained in an example, I'll give you one.

Ex. 1

$\int \frac {1}{x^2-4} \operatorname {d}x$

By doing some factoring we can see that $x^2 - 4$ also equals $(x+2)(x-2)$

I'm going to leave off the integral sign right now so you don't get confused and so you can see what I'm doing with the fraction

$\frac {A}{x + 2}+ \frac {B}{x - 2}$

In order to get the numerator we need to multiply by the LCD

$1=A(x-2) + B(x + 2)$

we set it equal to 1 because 1 is the number in the original numerator

Now we can choose to solve for A or B first. In order to do this we must find when one of them equals 0

when x = 2
$1 = A(0) + B(4)$

$B = \frac {1}{4}$

Now let's find A

When x = -2
$1 = A(-4) + B(0)$

$A = - \ frac{1}{4}$

This means that
$\frac {\left (- \frac {1}{4}\right)}{x-2} + \frac {\left (\frac {1}{4} \right)}{x+2}$

Now that we know what the partial fractions are we can now integrate

$\int \frac {1}{4(x+2)} \operatorname {d}x - \int \frac {1}{4(x-2)} \operatorname {d}x$

$= \frac {\ln(\left\vert {x+2}\right\vert)}{4} - \frac {\ln (\left\vert x-2 \right\vert)}{4}$

$= \frac {1}{4} \ln \left(\frac {\left\vert x+2 \right\vert}{\left\vert x-2 \right\vert}\right)$

Other problem to work on for practice

$\frac {x^2 + 3x + 2}{(x -2)^2 (x + 1)}$