My Math Forum  

Go Back   My Math Forum > High School Math Forum > Pre-Calculus

Pre-Calculus Pre-Calculus Math Forum


Thanks Tree8Thanks
Reply
 
LinkBack Thread Tools Display Modes
October 24th, 2018, 05:54 AM   #11
Senior Member
 
Joined: Oct 2009

Posts: 629
Thanks: 192

Quote:
Originally Posted by SDK View Post
To expand on what skipjack has done, EVERY polynomial of degree $d > 2$ can be factored into lower degree polynomials over the reals. Also EVERY polynomial of degree $d > 1$ can be factored into lower degree polynomials over the complex numbers. Hence, partial fractions works universally for problems like this.
Yes, in theory a problem like this is always solvable with partial fractions, but consider this:

$$\int \frac{1 - 5x^4}{(x^5 - x - 1)^2}dx$$

In theory it is indeed solvable with partial fractions. Maybe in practice too, I haven't tried it. But it's going to be very complicated because you can't find the explicit roots of the denominator. So I'm not at all sure if the method will work here.

However, an easy solution does exist here, a primitive is simply
$$\frac{1}{x^5 - x - 1}.$$
Thanks from happy21
Micrm@ss is offline  
 
October 25th, 2018, 12:02 PM   #12
Senior Member
 
happy21's Avatar
 
Joined: Jan 2012

Posts: 123
Thanks: 2

Quote:
Originally Posted by SDK View Post
To expand on what skipjack has done, EVERY polynomial of degree $d > 2$ can be factored into lower degree polynomials over the reals. Also EVERY polynomial of degree $d > 1$ can be factored into lower degree polynomials over the complex numbers. Hence, partial fractions works universally for problems like this.
True but the method of assuming Ax+B in the numerator works for only those polynomials in the denominator which have real rational factors. This I think is true..If its wrong, plz correct me. Thx.
happy21 is offline  
October 25th, 2018, 12:04 PM   #13
Senior Member
 
happy21's Avatar
 
Joined: Jan 2012

Posts: 123
Thanks: 2

Quote:
Originally Posted by romsek View Post
If I understand the problem... what you need to do is set up

$\dfrac{x^3-1}{(x^4+1)(x+1)} = \dfrac{A x^3 + B x^3 + C x + D}{x^4+1}+\dfrac{E}{x+1}$
Oh I see...Actually I wanted to know the numerator which is cubic with constants A, B, C & D. Thanks. This is to be assumed.
happy21 is offline  
Reply

  My Math Forum > High School Math Forum > Pre-Calculus

Tags
integral, plz, solve



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
How can I solve such an Integral? ghafarimahsa Calculus 1 September 11th, 2013 01:07 PM
How to solve this integral? kaspersky0 Complex Analysis 3 October 15th, 2012 10:19 AM
How to solve an integral. Ad van der ven Calculus 3 December 17th, 2011 10:53 AM
An integral to solve malaguena Calculus 0 February 14th, 2011 02:18 AM
please solve this integral ..thx art2 Calculus 1 September 24th, 2009 05:46 AM





Copyright © 2018 My Math Forum. All rights reserved.