1. ## Integration.

Hello i have a question pplz help me it is
how the:
a is a constant number

after integration became

Thanks

2. Because your are integrating with respect to the variable p, as that variable is changing not a, thats the way I see it

3. ## Hello

$\displaystyle P^2+1=U$
and finding:
$\displaystyle p=sqrt(U-1)$
and findinf dp and subsitituing them into the original equation etc....
if there is easier method plz give it to me
Thnx

4. I am working on your problem as well, I am still a learner compared to everyone here, I have tried you substitution, before you mention it as it gets tad confusing im re-working it, hope I can help you

5. ## ok

i dunno i have tried many other method and didnt work
Thanks Anyway

6. This problem is very drawn out for me, I will find out how to do it though! It will be good for both of us, but our methods seem to complicate thing while integrating and simplifying as I told by my professor, mathematicians aim to make hard things very simple when working out a problem, so we will get it! I am sure there a few around here that this is easy for, I was close to the solution but made a careless everyone and wound up with a few missing parts, like the numerator coming to p^2 instead of p^3

Lets see if we can make sense of things together, first we know -3a is a constant so we can pull that to the outside and that the numerator $\displaystyle p^2$ when integrated as if it was just alone as such$\displaystyle p^2$ becomes $\displaystyle \frac{p^3}{3}$ and remember we pulled out the -3a so now the the integrated form we get $\displaystyle -3a*\frac{p^3}{3}$ we get $\displaystyle -a * p^3$ so somehow we need to make a relation to the denominator and decide the right method of integrating to where we can relate to this

Edit: have you tried integration by parts, I feel I am close using this method

7. Have you tried the substitution $\displaystyle p = \frac{1}{x}$ ?

since

$\displaystyle dp = \frac{-1}{x^2} dx$

the integral ( let's omit the constant $\displaystyle -3a$) becomes

$\displaystyle \int \frac{x^5}{\sqrt{x^2+1}^5} \cdot \frac{1}{x^2} \cdot \frac{-1}{x^2} dx$

$\displaystyle = - \int \frac{x dx}{\sqrt{x^2+1}^5}$

$\displaystyle = - (1 + x^2)^{\frac{-3}{2}} (\frac{-2}{3}) \frac{1}{2} + C$

$\displaystyle = \frac{1}{3} \frac{ 1}{ \sqrt{ 1 + \frac{1}{p^2} }^3 } + C$

$\displaystyle = \frac{ p^3}{3 \sqrt{ 1 + p^2}^3 } + C$

8. Jeepers, the one I never thought of trying, how did you know what to correctly substitute?
Also where x^5 comes from?

9. Originally Posted by RockHard
Jeepers, the one I never thought of trying, how did you know what to correctly substitute?

If thec0o0lest didn't give us the answer , i won't think of this substitution .

I observed that the integral is algebraic function , not transcendental ( usually $\displaystyle \sinh^{-1}(x) , \cosh^{-1}(x)$ ) .

so i attempt to change the variable ( algebraic ) . It is just a trial !

Another case we can use this sub. is

$\displaystyle \int \frac{dx}{\sqrt{x^2+1}^3}$

to

$\displaystyle - \int \frac{u ~du}{\sqrt{u^2+1}^3}$

10. simplependulum Thankssssssssssssssssssssssssss

11. Originally Posted by RockHard
Also where x^5 comes from?
$\displaystyle x^5$ comes from the denominator ( after changing $\displaystyle p$ to $\displaystyle \frac{1}{x}$)

$\displaystyle \frac{1}{\sqrt{p^2+1}^5} = \frac{1}{\sqrt{1 + \frac{1}{x^2}}^5 }$

12. Originally Posted by RockHard
Jeepers, the one I never thought of trying, how did you know what to correctly substitute?
Also where x^5 comes from?
the x^5 come from
$\displaystyle sqrt{1/x^2+1}^5$
when you subtitute p=1/x
you need to make some balance in the denominator you multiply by

$\displaystyle x^(5/2)$

13. Man, this is a confusing integral, I was stuck on it, If I did not see the answer I wouldn't even came to close to guessing a right answer