Simplifying Complicated Fractions for Easier Calculus Differentiation

[ggolub]

Hello folks, I am studying Calculus 1 by S. Tohmpson 'Calculus made easy'.
On p. 130 I am stuck with few quizes which require to split into fractions:

(3x^2+2x+1)/[(x+2)(x^2+x+1)]
and
(x^4+1)/(x^3+1)

It asks to break each expression into sum of more simple fractions so following it differentiation would be much simpler.
Please let me know if you know how to do it.

 

[HallsofIvy]

Hello folks, I am studying Calculus 1 by S. Tohmpson 'Calculus made easy'.
On p. 130 I am stuck with few quizes which require to split into fractions:

(3x^2+2x+1)/[(x+2)(x^2+x+1)]

Since the denominator has higher degree than the numerator and x^2+ x+ 1 cannot be factored in terms of real numbers, this can be written
$$\frac{A}{x+ 2}+ \frac{Bx+ C}{x^2+ x+ 1}$$
On method of doing that is write
$$\frac{3x^2+ 2x+ 1}{(x+1)(x^2+ x+ 1)}= \frac{A}{x+ 2}+ \frac{Bx+ C}{x^2+ x+ 1}$$
multiply both sides by $(x+1)(x^2+ x+ 1)$ to eliminate the fractions:
[tex] 3x^2+ 2x+ 1= A(x^2+ x+ 1)+ (Bx+ C)(x+ 2)[/itex]
Since this is to be true for all x, you can let x be any 3 numbers you want to get 3 equations to solve for A, B,a nd C. For example x=-2 is especialy nice since then you get $$3(-2)^2+ 2(-2)+ 1= 12- 4+ 1= 9= A((-2)^2+ (-2)+ 1)= 3A$$
But taking x= 0 gives
$$1= A+ 2C$$
and x= 1 gives
$$6= 3A+ 3B+ 3C$$

and
(x^4+1)/(x^3+1)

Here the numerator has higher degree than the denominator so the first thing you want to do is use "long division". $x^3+ 1$ divides into $x^4+ 1$ x times with -x+ 1 as remainder:
$$\frac{x^4+ 1}{x^3+ 1}= x+ \frac{1- x}{x^3+ 1}= x+ \frac{1- x}{(x- 1)(x^2+ x+ 1)}$$
Now find A, B, and C so that
$$\frac{1- x}{x^2+ x+ 1}= \frac{A}{x-1}+ \frac{B}{x^2+ x+ 1}$$.

It asks to break each expression into sum of more simple fractions so following it differentiation would be much simpler.
Please let me know if you know how to do it.

Since this has nothing to do with "Differential Equations", I am moving it to "Calculus and Analysis".

 

[ggolub]

Thank you so much!
Can you please refer me the book or name of topic which I can use to find this material. I want to understand the method behind it.

 

[Strants]

$$(x-1)(x^{2}+x+1) = x^{3} - 1$$, not $$x^{3} + 1$$.

Cubics are way beyond my ability to factor, but just looking at it, but I'm pretty sure $$x^{3} + 1$$ is in lowest terms. In that case, wouldn't we leave the equation in the form it is already in? Or am I missing something?

 

[HallsofIvy]

You are right. $(x- 1)(x^2+ x+ 1)= x^3- 1$.

What I meant to say was $x^3+ 1= (x+ 1)(x^2- x+ 1)$

$x^n+ 1$ cannot be factored (in terms of real numbers) if n is even but if n is odd, $x^n+ 1= (x- 1)(x^{n-1}- x^{n-2}+ \cdot\cdot\cdot- x+ 1)$.