Integration

SoSolid · 5 Nov 2006 at 14:59

Ok guys, I am finding this topic hard for some reason. How do you integrate the following question: x+1 / sq root (x^2 +2x+5) when the limits are 1 and 0.

Van Hellseek · 5 Nov 2006 at 14:59

i thought this was going to be yet another thread about muslims.

SoSolid · 5 Nov 2006 at 15:12

lol that made me laugh!

qwerty · 5 Nov 2006 at 15:21

How do you integrate a muslim?

Take off their veil!

SoSolid · 5 Nov 2006 at 15:27

oh please as much as these are making me laugh I am so stuck on this question, Arcade fire where are you my friend....please please help me.

dirtydog · 5 Nov 2006 at 16:11

/enters thread ready to talk about immigrants

/leaves feeling disappointed

Hamzter · 5 Nov 2006 at 16:41

That's quite a tough one, I thiiiink you can do it with Integration by Parts, but you'll have to give me a few mins. What level is this at anyway?

SoSolid · 5 Nov 2006 at 16:42

First year degree, I have forgot all my A level shizzle when it comes to integration.

SoSolid · 5 Nov 2006 at 16:55

The answer I got was: 2ln(x^2 + 2x + 5)^-1/2 + c . I do not know if I am correct in terms of what I have done with the square root?

Hamzter · 5 Nov 2006 at 16:58

Ok that's got me stumped, it can't be done by parts cos there'll still be an integral at the end of it, it's been a while since I did any tricky integration, havn't done any since first year of uni (now in 3). I'll keep thinking...

Edit: Just wondering, is that x+1 meant to be in brackets? or is it x + (1/sqrt(x^2+2x+5))?

SoSolid · 5 Nov 2006 at 16:59

Its got to be done by substitution as it is a function/function.

titaniumx3 · 5 Nov 2006 at 17:08

Yep, use a subtitution, say u=x^2+2*x+5.

Should end up with the answer: 2*sqrt(2)-sqrt(5)

Hamzter · 5 Nov 2006 at 17:20

^^^ What he said.

use substitution u = x^2 + 2x + 5 so du/dx = 2x + 2 your equation simplifies to

integral(1/(2(u^0.5)))du

integrate that and u get [u^1/2], giving 8^0.5-5^0.5.

Psyk · 5 Nov 2006 at 17:21

I seem to remember the completing the square method from A-level but I was never much good at it. Pretty sure it could be used to integrate equations of that form.

Kainz · 5 Nov 2006 at 17:21

seek said:
i thought this was going to be yet another thread about muslims.

lol ditto /me leaves...

triggerthat · 5 Nov 2006 at 17:59

The answer is infinity.

Psyk · 5 Nov 2006 at 17:59

qwerty said:
How do you integrate a muslim?

Van Hellseek · 5 Nov 2006 at 18:06

DaveF · 6 Nov 2006 at 11:28

Psyk said:
I seem to remember the completing the square method from A-level but I was never much good at it. Pretty sure it could be used to integrate equations of that form.

It's not really going to help here - the point is the "obvious" substitution has the same derivative as the top bit of the fraction, which makes doing it that way easy-peasy.

Arcade Fire · 6 Nov 2006 at 12:42

SoSolid said:
The answer I got was: 2ln(x^2 + 2x + 5)^-1/2 + c . I do not know if I am correct in terms of what I have done with the square root?

Nairn, you shouldn't be getting logarithms here. You only get those if the top of the fraction is the derivative of the bottom, which isn't the case here because of the square root.

In this case you want to use a substitution, as has been said. There are two obvious choices, which are u = x^2 + 2x + 5, or alternatively u = (x^2 + 2x + 5)^{1/2}.

Doing either of these will give you a nice easy integral in u to perform. Hamzter did the first one correctly. Using the second substitution instead is in some ways even nicer, since you'll end up with

du/dx = (x+1)/u

so that when you write 'du dx/du' to replace the 'dx' you'll find that everything cancels, and you just have to do

Int 1 du, with the limits sqrt(8) and sqrt(5).