My Math Forum Are there issues with this?

 Calculus Calculus Math Forum

 April 21st, 2012, 08:50 PM #1 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Are there issues with this? \begin{align} \int \frac{x^2-3}{x^4+3}\,dx &= \int\frac{x^2}{x^4+3}\,dx-3\int\frac{1}{x^4+3}\,dx x^4+3 &= (x+t)^2(x-it)^2 \quad \text{N.B. } t=(-3)^{\frac{1}{4}}\\ \int \frac{x^2}{x^4+3} &= A\int \frac{1}{(x+t)^2}\,dx+B\int \frac{1}{(x-it)^2} \, dx\\ A&=\frac{t^2}{(-t-it)^2}\\ B&=\frac{-t^2}{(t+it)^2}\\ \int \frac{x^2}{x^4+3}\,dx &= \frac{t^2}{(-t-it)^2}\left(-\frac{1}{x+t}\right)+\frac{-t^2}{(t+it)^2}\left(-\frac{1}{x-it}\right)\\ &=\frac{t^2}{(t+it)^2(x-it)}-\frac{t^2}{(-t-it)^2(x+t)}\\ \int \frac{1}{x^4+3}\,dx &= A'\int \frac{1}{(x+t)^2}\,dx+B'\int \frac{1}{(x-it)^2}\,dx\\ A'&=\frac{1}{(-t-it)^2}\\ B'&=\frac{1}{(t+it)^2}\\ \int \frac{1}{x^4+3}\,dx &=\frac{-1}{(-t-it)^2(x+t)}+\frac{-1}{(t+it)^2(x-it)}\\ \int \frac{x^2-3}{x^4+3}\,dx &=\frac{t^2+3}{(t+it)^2(x-it)}-\frac{t^2+3}{(-t-it)^2(x+t)}\\ &=\frac{t^2+3}{(t+it)^2(x-it)}-\frac{t^2+3}{(t+it)^2(x+t)} \end{align}
 April 21st, 2012, 09:12 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Are there issues with this? I simplified your result some, then differentiated and did not get back the original integrand. That is a real bear of an integral , at least by factoring the denominator: $x^4+3=$$x^2+\sqrt{2}\sqrt[4]{3}x+\sqrt{3}$$$$x^2-\sqrt{2}\sqrt[4]{3}x+\sqrt{3}$$$ and then using partial fraction decomposition, substitutions, etc.
April 21st, 2012, 10:04 PM   #3
Senior Member

Joined: Jul 2011

Posts: 245
Thanks: 0

Re: Are there issues with this?

Quote:
 Originally Posted by MarkFL I simplified your result some, then differentiated and did not get back the original integrand. That is a real bear of an integral
I'm aware. That is rather crazy. Can you post your work? Also, is there an error algebraically in what I wrote? It wasn't really easy to compose, but the idea seemed simple.

EDIT: I made a stupid mistake... Namely,

$x^4+3=(x-it)(x+it)(x+t)(x-t)\neq(x-it)^2(x+t)^2$

Redoing and reposting, one moment. . .

 April 21st, 2012, 10:14 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Are there issues with this? I simply looked at the suggested method given by wolframalpha.com. Your factorization of $x^4+3$...nevermind, I see you caught it.
 April 21st, 2012, 10:30 PM #5 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Re: Are there issues with this? \begin{align} \int \frac{x^2}{x^4+3}\,dx&=A\int\frac{1}{x-it}\,dx+B\int\frac{1}{x+it}\,dx+C\int\frac{1}{x+t} \,dx+D\int\frac{1}{x-t}\,dx\\ A&=\frac{-t^2}{(2it)(it+t)(it-t)}\\ B&=\frac{-t^2}{(-2it)(-it+t)(-it-t)}\\ C&=\frac{t^2}{(-t-it)(-t+it)(-2t)}\\ D&=\frac{t^2}{(t-it)(t+it)(2t)}\\ \int \frac{x^2}{x^4+3}\,dx&=A\frac{-1}{x-it}+B\frac{-1}{x+it}+C\frac{-1}{x+t}+D\frac{-1}{x-t}\\ \int \frac{1}{x^4+3}\,dx&=A'\int\frac{1}{x-it}\,dx+B'\int\frac{1}{x+it}\,dx+C'\int\fr ac{1}{x+t}\,dx+D'\int\frac{1}{x-t}\,dx\\ A'&=\frac{1}{(2it)(it+t)(it-t)}\\ B'&=\frac{1}{(-2it)(-it+t)(-it-t)}\\ C'&=\frac{1}{(-t-it)(-t+it)(-2t)}\\ D'&=\frac{1}{(t-it)(t+it)(2t)}\\ \int \frac{1}{x^4+3}\,dx&=A'\frac{-1}{x-it}+B'\frac{-1}{x+it}+C'\frac{-1}{x+t}+D'\frac{-1}{x-t}\\ \int \frac{x^2-3}{x^4+3}\,dx&=\frac{3A'-A}{x-it}+\frac{3B'-B}{x+it}+\frac{3C'-C}{x+t}+\frac{3D'-D}{x-t}\\ \end{align} How's this?
 April 21st, 2012, 10:37 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Are there issues with this? You don't seem to actually be integrating...you should have natural logarithms in your result, i.e.,: $\int\frac{1}{x+k}\,dx=\ln|x+k|+C$ where $k\in\mathbb R$
 April 21st, 2012, 10:41 PM #7 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Re: Are there issues with this? Woooooooow. -facepalm- This is why you never try to integrate and multitask. It works out to, instead, $\int \frac{x^3-3}{x^4+3}\,dx=(A-3A#39\ln|x-it|+(B-3B'\ln|x+it|+(C-3C'\ln|x+t|+(D-3D'\ln|x-t|+C" /> Thank you for the patience, Mark. The nuances and little things are what I always miss. EDIT: The big reason I asked this topic was because I was uncertain if extending the method of partial fraction decomposition to complex numbers was a bad idea. Wolfram Alpha does not seem to like it, but it makes sense to me.
 April 21st, 2012, 10:52 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Are there issues with this? Another issue: You cannot state: $\frac{1}{(x-a)(x-b)}=\frac{1}{x-a}+\frac{1}{x-b}$ You need to use partial fraction decomposition. For example: $\frac{1}{(x-a)(x-b)}=\frac{1}{(a-b)(x-a)}+\frac{1}{(b-a)(x-b)}$ With 4 linear factors in the denominator it will be more complicated. You want: $\frac{1}{(x+it)(x-it)(x+t)(x-t)}=\frac{1}{4t^3}$$\frac{1}{x-t}-\frac{1}{x+t}+\frac{i}{x-it}-\frac{i}{x+it}$$$
April 21st, 2012, 11:01 PM   #9
Senior Member

Joined: Jul 2011

Posts: 245
Thanks: 0

Re: Are there issues with this?

Quote:
 Originally Posted by MarkFL Another issue: You cannot state: $\frac{1}{(x-a)(x-b)}=\frac{1}{x-a}+\frac{1}{x-b}$ You need to use partial fraction decomposition. For example: $\frac{1}{(x-a)(x-b)}=\frac{1}{(a-b)(x-a)}+\frac{1}{(b-a)(x-b)}$ With 4 linear factors in the denominator it will be more complicated. You want: $\frac{1}{(x+it)(x-it)(x+t)(x-t)}=\frac{1}{4t^3}$$\frac{1}{x-t}-\frac{1}{x+t}+\frac{i}{x-it}-\frac{i}{x+it}$$$
I don't understand this issue. I am using partial fraction decomposition...

Here are the steps:

$\frac{x^2}{x^4+3}=\frac{A}{x-it}+\frac{B}{x+it}+\frac{C}{x+t}+\frac{D}{x-t}$

This gives us an equation:

$A(x+it)(x+t)(x-t)+B(x-it)(x+t)(x-t)+C(x-it)(x+it)(x-t)+D(x-it)(x+it)(x+t)=x^2$

I evaluated this equation at the zeros. (it, -it, t, -t) This gave me the expressions for A, B, C, and D. Take A as an example:

$x=it \Rightarrow A(it+it)(it+t)(it-t)=(it)^2 \Rightarrow A=\frac{-t^2}{(2it)(it+t)(it-t)}$

 April 21st, 2012, 11:15 PM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Are there issues with this? My apologies, I see that now. You are even using the cover-up method. You can do a lot of simplification of your constants. For example: $A=\frac{-t^2}{(2it)(it+t)(it-t)}=-\frac{1}{4it}$ I find that that coefficient should be (using W|A): $A=-\frac{i}{4t}$ which is simply the negative of what you have.

 Tags issues

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Brightstar Geometry 3 January 19th, 2012 01:39 PM tinyone Calculus 5 September 11th, 2010 09:00 AM raiseit Math Events 1 June 1st, 2010 02:38 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top