I'm stumped on these:

The integral of x^3 (1+x^2)^3/2

The integral of the square root of (x^2 - 1)dx/x^3

The integral of x^3 dx/(4-x^2)^2

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- Feb 12th 2009, 09:49 PMGreySweatshirt306A Few Integrals
I'm stumped on these:

The integral of x^3 (1+x^2)^3/2

The integral of the square root of (x^2 - 1)dx/x^3

The integral of x^3 dx/(4-x^2)^2 - Feb 12th 2009, 09:53 PMJhevon
do a substitution, $\displaystyle u = 1 + x^2$

Quote:

The integral of the square root of (x^2 - 1)dx/x^3

Quote:

The integral of x^3 dx/(4-x^2)^2

- Feb 12th 2009, 10:04 PMGreySweatshirt306
Thanks for the help--but I don't see how a substitution works for either of those two problems.

- Feb 12th 2009, 10:39 PMJhevon
did you try it?

$\displaystyle \int {x^3}{(1 + x^2)^{3/2}}~dx$

Let $\displaystyle u = 1 + x^2 \implies \boxed{x^2 = u - 1}$ ....note that $\displaystyle x^3 = x \cdot x^2$

$\displaystyle \Rightarrow du = 2x~dx$

$\displaystyle \Rightarrow \frac 12 ~du = x~dx$

Thus our integral becomes

$\displaystyle \frac 12 \int \frac {u - 1}{u^{3/2}}~du$

i suppose you can take it from here. the other integral is similar - Feb 17th 2009, 06:45 AMGreySweatshirt306Trig Substitution
Is there anyway to do these using a trig sub?

- Feb 17th 2009, 07:15 AMKrizalid
Utterly unnecessary.

1) $\displaystyle x=\tan\varphi.$

2) $\displaystyle x=\sec\varphi.$

3) $\displaystyle x=2\sin\varphi.$ - Feb 17th 2009, 09:47 AMGreySweatshirt306Thanks
Thanks for the help. I'm not very good at these though, could you flesh them out a bit? I really appreciate anything you can do

- Feb 17th 2009, 04:45 PMJhevon
- Feb 17th 2009, 05:35 PMGreySweatshirt306Yeah
Trig Subs are required

- Feb 17th 2009, 05:37 PMKrizalid
You didn't specify that from the beginning, and that's bad, pretty bad since you wasted Jhevon's time.

If you want to be helped, show at least what you've tried. - Feb 17th 2009, 05:53 PMGreySweatshirt306
Well let me edit that--

Trig subs are NOT required, what I meant to say is that I came to my math teacher with some of the work that Jhevon got me started on (thanks again), and he instructed me to also try to work them out using a trig substitution, so i'm back here. Sorry for the confusion