1. ## A 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

2. Originally Posted by GreySweatshirt306
I'm stumped on these:

The integral of x^3 (1+x^2)^3/2
do a substitution, $u = 1 + x^2$

The integral of the square root of (x^2 - 1)dx/x^3
note that $\frac {x^2 - 1}{x^3} = \frac {x^2}{x^3} - \frac 1{x^3} = \frac 1x - x^{-3}$

The integral of x^3 dx/(4-x^2)^2
another substitution problem, let $u = 4 - x^2$

3. Thanks for the help--but I don't see how a substitution works for either of those two problems.

4. Originally Posted by GreySweatshirt306
Thanks for the help--but I don't see how a substitution works for either of those two problems.
did you try it?

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

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

$\Rightarrow du = 2x~dx$

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

Thus our integral becomes

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

i suppose you can take it from here. the other integral is similar

5. ## Trig Substitution

Is there anyway to do these using a trig sub?

6. Utterly unnecessary.

1) $x=\tan\varphi.$

2) $x=\sec\varphi.$

3) $x=2\sin\varphi.$

7. ## Thanks

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

8. Originally Posted by GreySweatshirt306
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
what's wrong with the way they were done before? are you required to use trig subs?

9. ## Yeah

Trig Subs are required

10. 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.

11. 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