### Partial fractions

Now we return to a more special but still important technique of doing
indefinite integrals. This depends on a good trick from algebra to
transform complicated *rational functions* into simpler
ones. Rather than try to formally describe the general fact, we'll do
the two simplest families of examples.

Consider the integral $$\int {1\over x(x-1)}\; dx$$ As it stands, we do not recognize this as the derivative of anything. However, we have $${1\over x-1}-{1\over x}={x-(x-1)\over x(x-1)}={1\over x(x-1)}$$ Therefore, $$\int {1\over x(x-1)}\; dx=\int {1\over x-1}-{1\over x}\;dx=\ln(x-1)-\ln x + C$$ That is, by separating the fraction $1/x(x-1)$ into the ‘partial’ fractions $1/x$ and $1/(x-1)$ we were able to do the integrals immediately by using the logarithm. How to see such identities?

Well, let's look at a situation
$${cx+d\over (x-a)(x-b)}={A\over x-a}+{B\over x-b}$$
where $a,b$ are given numbers (not equal) and we are to *find*
$A,B$ which make this true. If we can find the $A,B$ then we can
integrate $(cx+d)/(x-a)(x-b)$ simply by using logarithms:
$$\int {cx+d\over (x-a)(x-b)}\;dx=\int{A\over x-a}+{B\over x-b}\;dx=
A\ln(x-a)+B\ln(x-b)+C$$
To find the $A,B$, multiply through by $(x-a)(x-b)$ to get
$$cx+d=A(x-b)+B(x-a)$$
When $x=a$ the $x-a$ factor is $0$, so this equation becomes
$$c\cdot a+d=A(a-b)$$
Likewise, when $x=b$ the $x-b$ factor is $0$, so we also have
$$c\cdot b+d=B(b-a)$$
That is,
$$A={c\cdot a+d\over a-b}\;\;\;\;\;B={c\cdot b+d\over b-a}$$
So, yes, we can find the constants to break the fraction
$(cx+d)/(x-a)(x-b)$ down into simpler ‘partial’ fractions.

*Further*, if the numerator is of *bigger degree*
than $1$, then before executing the previous algebra trick we must
first *divide the numerator by the denominator to get a remainder
of smaller degree*. A simple example is
$${x^3+4x^2-x+1\over x(x-1)}=?$$ *We must recall how to divide
polynomials by polynomials and get a remainder of lower degree than
the divisor*. Here we would divide the $x^3+4x^2-x+1$ by
$x(x-1)=x^2-x$ to get a remainder of degree less than $2$ (the degree
of $x^2-x$). We would obtain
$${x^3+4x^2-x+1\over x(x-1)}=x+5+{4x+1\over x(x-1)}$$ since the
quotient is $x+5$ and the remainder is $4x+1$. Thus, in this situation
$$\int {x^3+4x^2-x+1\over x(x-1)}\;dx=\int
x+5+{4x+1\over x(x-1)}\;dx$$
Now we are ready to continue with the *first* algebra trick.

In this case, the first trick is applied to $${4x+1\over x(x-1)}$$ We want constants $A,B$ so that $${4x+1\over x(x-1)}={A\over x}+{B\over x-1}$$ As above, multiply through by $x(x-1)$ to get $$4x+1=A(x-1)+Bx$$ and plug in the two values $0,1$ to get $$4\cdot 0+1=-A\;\;\;\;\;4\cdot 1+1=B$$ That is, $A=-1$ and $B=5$.

Putting this together, we have $${x^3+4x^2-x+1\over x(x-1)}=x+5+{-1\over x}+{5\over x-1}$$ Thus, $$\int {x^3+4x^2-x+1\over x(x-1)}\;dx=\int x+5+{-1\over x}+{5\over x-1}\;dx$$ $$={x^2\over 2}+5x-\ln x+5\ln(x-1)+C$$

In a slightly different direction: we can do any integral of the form $$\int {ax+b\over 1+x^2}\;dx$$ because we know two different sorts of integrals with that same denominator: $$\int{1\over 1+x^2}\;dx=\arctan x+C\;\;\;\;\; \int {2x\over 1+x^2}\;dx=\ln(1+x^2)+C$$ where in the second one we use a substitution. Thus, we have to break the given integral into two parts to do it: $$\int {ax+b\over 1+x^2}\;dx={a\over 2}\int {2x\over 1+x^2}\;dx +b\int{1\over 1+x^2}\;dx$$ $$={a\over 2}\ln(1+x^2)+b\arctan x+C$$

And, as in the first example, if we are given a numerator of
degree $2$ or larger, then we *divide* first, to get a remainder
of lower degree. For example, in the case of
$$\int {x^4+2x^3+x^2+3x+1\over 1+x^2}\;dx$$
we divide the numerator by the denominator, to allow us to write
$${x^4+2x^3+x^2+3x+1\over 1+x^2}=x^2+2x+{x+1\over 1+x^2}$$
since the quotient is $x^2+2x$ and the remainder is $x+1$. Then
$$\int {x^4+2x^3+x^2+3x+1\over 1+x^2}\;dx=
\int x^2+2x+{x+1\over 1+x^2}$$
$$={x^3\over 3}+x^2+{1\over 2}\ln(1+x^2)+\arctan x+C$$

These two examples are just the simplest, but illustrate the idea of using algebra to simplify rational functions.

#### Exercises

- $\int { 1 \over x(x-1) }\,dx=?$
- $\int { 1+x \over 1+x^2 }\,dx=?$
- $\int { 2x^3+4 \over x(x+1) }\,dx=?$
- $\int { 2+2x+x^2 \over 1+x^2 }\,dx=?$
- $\int { 2x^3+4 \over x^2-1 }\,dx=?$
- $\int { 2+3x \over 1+x^2 }\,dx=?$
- $\int { x^3+1 \over (x-1)(x-2) }\,dx=?$
- $\int { x^3+1 \over x^2+1 }\,dx=?$

#### Thread navigation

##### Calculus Refresher

- Previous: Integration by parts
- Next: Trigonometric integrals