PDA

View Full Version : More integration



qwertysaur
03-23-2009, 05:53 PM
I have a very annoying problem that I can't get. =/

∫ (X + 1)/(x<sup>2</sup> + 4x + 8) dx

Aerith's Knight
03-26-2009, 07:16 PM
I just filled two pages with partial integrals, imaginary numbers, splitting equations and a whole bunch of other crap. Then I decided to look at the answer, which came to

-0.5 * ArcTan[(2+x)/2] + 0.5*Log[x^2 +4x +8]

Which will be annoyingly complicated. So could you check first if the question itself wasn't wrong?

I can get you started by:

∫ (X + 1)/(x2 + 4x + 8) dx = ∫ X/(x2 + 4x + 8) dx + ∫ 1/(x2 + 4x + 8) dx

Where ∫ 1/(x2 + 4x + 8) dx = ln(x2 + 4x + 8)/(2x + 4) + C

Also, x2 + 4x + 8 = (x - 2 - 2j)(x - 2 + 2j), but I doubt that helps.

qwertysaur
03-30-2009, 05:59 PM
I feel bad that you had two pages of work to do, but I really appreciate it. I took the derivative of your answer, and now I get it. Thanks :)

SeeDRankLou
03-30-2009, 08:25 PM
Is the "X" a typo for "x", or is the "X" different than the "x"?

rubah
03-30-2009, 11:31 PM
Sepho felt that he must share his solution with you all
http://rubah.net/sepho/solution.png

Jessweeee♪
04-06-2009, 05:42 PM
The day is saved!