# Partial fractions, chain rule

#### EvLer

Hi, I have 2 questions:

1. partial fractions:
if I have following integral: Itegral[(1-2x^2)/(x - x^3)]dx;
my question is do I break down the denominator to x(1-x^2) or do I go further:
x(1-x)(1+x); this way it becomes more complicated;

2. chain rule:
how does chain rule work in this case: f(x)g(x)h(x)?
is it something like f'(x)[g(x)h(x)] + g'(x)[f(x)h(x)] + h'(x)[f(x)g(x)] ??? or how would I differentiate this?

Related Introductory Physics Homework News on Phys.org

#### StatusX

Homework Helper
1. You need enough unknowns to be able to equate the three coefficients for each power of x in the numerator (1,0,-2). So you'll need to break it up into all three factors. In some cases (not this one) the two factors alone would work, but only if the numerator happens to be just right. It would be like finding the point where two lines intersect in 3D; you can only find it if the lines happen to intersect.

2. Think of it as (fgh)' = f'(gh) + f(gh)'. Your guess is right.

#### Yegor

In first 1-2x^2=1-x^2-x^2 is very helpful. Then you have 2 very easy integrals

#### EvLer

Aaaah, i see. Thanks for help.

#### OlderDan

Homework Helper
#2 is not the chain rule. That is the product rule. It does work as you have it

#### EvLer

OlderDan said:
#2 is not the chain rule. That is the product rule. It does work as you have it
Right, sorry, mixed up the names
Chain rule is differentiating what I call "embedded functions".

#### whozum

EvLer said:
Right, sorry, mixed up the names
Chain rule is differentiating what I call "embedded functions".
Composite functions is their real name, btw.

#### EvLer

*mental note to self* TERMINOLOGY!
I knew it, but did not recall on the spot, thanks everyone!

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving