Solving integrals using partial fractions

Question

I'm learning partial fractions. However, I am not sure about when to put $A, B, C..$ into the numerator and when to put $Ax+B, Bx+C..$ into the numerator.

For example, I have the following integral:

$$\int \frac{x^5}{(x-1)^2(x^2-1)}dx$$

and I want to calculate simplify using partial fractions.
I know that $(x-1)^2$ has just one real solution, so I will write it as $\frac{A}{x-1}+\frac{B}{(x-1)^2}$.
However, $x^2-1$ has two real solutions, $\{-1, 1\}$.
Do I write is as $\frac{Cx+D}{(x^2-1)}$ or just $\frac{C}{(x^2-1)}$?

I am really not sure about when to use the simple form $A,B$ and when to use $Cx+D, Dx+E$.

Thanks

Answer 1

Actually, since the degree of the numerator is not smaller than the degree of the denominator, your first step should be to write $x^5$ as$$(x+2)\times\bigl((x-1)^2(x^2-1)\bigr)+4x^3-2x^2-3x+2.$$So$$\frac{x^5}{(x-1)^2(x^2-1)}=x+2+\frac{4x^3-2x^2-3 x+2}{(x-1)^2(x^2-1)}.$$On the other hand,$$x^2-1=(x-1)(x+1)\implies(x-1)^2(x^2-1)=(x-1)^3(x+1)$$and therefore you should try to get $A$, $B$, $C$ and $D$ such that$$\frac{4x^3-2x^2-3 x+2}{(x-1)^2(x^2-1)}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{(x-1)^3}+\frac D{x+1}.$$

Answer 2

A general rule of thumb is to never leave things that can be further simplified. For example,

• prefer $\frac{A}{x+1} + \frac{B}{x-1}$ to $\frac{Cx+D}{x^2-1}$

• prefer $\frac{A}{x}+\frac{B}{x^2}$ to $\frac{Ax+B}{x^2}$

Of course, you are not obliged to follow this rule to the point - if you see a clear gain to not use, do not use it.