6.3.2 Finding an Antidifferential

The first thoughts that usually appear in response to

"The answer is `1/(xtext[(]c-x text[)])`; what is the question?"

are

1. Multiplication of algebraic fractions has answers that look like this.
2. Addition of algebraic fractions has answers that look like this.

If multiplication is what we are looking for, finding the direct problem is easy:

That's somewhat promising, because each of the factors on the left is a reciprocal [`text[(]-1text[)]`-th power] of a linear expression in `x`. However, simple products don't turn up that often in derivatives — except in the Chain Rule — and the left side of the last equation doesn't look like a Chain Rule calculation. The Product Rule leads to a sum of products, and we don't have that either. Thus, while

is a correct way to rewrite the function we want to antidifferentiate, it doesn't seem to lead to a solution.

Addition of fractions may be more useful for purposes of antidifferentiation because the Sum Rule for derivatives says the right thing: The derivative of a sum is the sum of the derivatives. As we have seen, it follows immediately that the same is true for antiderivatives. However, the inverse algebraic problem is not quite so obvious as was

We want something like

where the numerators `A` and `B` are yet to be determined.

Activity 2

1. Find numbers `A` and `B` so that

   ${\displaystyle \frac{A}{x}+\frac{B}{2-x}=\frac{1}{x(2-x)}\mathrm{.}}$

2. Repeat the calculation in (a) with `2` replaced by an arbitrary constant `c`, i.e., find numbers `A` and `B` so that

   ${\displaystyle \frac{A}{x}+\frac{B}{c-x}=\frac{1}{x(c-x)}}$.

   (This time your expressions for `A` and `B` will involve `c`.)

Comment on Activity 2

Checkpoint 1

Now we are ready to solve the antidifferentiation problem. Using the Sum and Constant Multiple Rules, we can write

An antidifferential of `(dx)/(xtext[(]c-x text[)])`

= an antidifferential of `(1/c 1/x+1/c 1/(c-x))dx`

= the sum of antidifferentials of `1/c (dx)/x` and `1/c (dx)/(c-x).`

Activity 3

Explain why, when `x` is restricted to the interval between `0` and `c`, an antidifferential of `(dx)/[xtext[(]c-x text[)]]` is

Comment on Activity 3

Checkpoint 2