Question

integral problem

Answer 1

integral of 1/sqrt(x^2+6x)dx

Answer 2

Are you actually typing anything mathmale? XD

Answer 3

I also need help with integral x^2*sqrt(1+x)

Answer 4

There are various ways in which to integrate here. The first methods that come to mind are (1) completing the square, followed by substitution, and (2) trignonometric substitution.

completing the square is probably the easier.

x^2+6x has the makings of a perfect square. I'd prefer to assume you already know something ab out completing the square, so will just go through the steps, while encouraging you to ask questions if the process is not clear.

x^2 + 6x +(3)^2 - (3)^2 is exactly the same thing as x^2+6x.
But:
x^2 + 6x + 9 - 9
reduces to the simpler

(x-3)^2 -9.


Then y our ingral is

dx
Int ( -----------------)
Sqrt( (x-3)^2 - 9)

this could be integrated easily after a substitution. Hint: let u=x-3 and find du, then

du
evaluate Int ( ------------- )
Sqrt (u^2 - 9)

Unfortunately, I think we'd need a trig subst. here.

Answer 5

Still, the problem is now easier than it was before.

Answer 6

Personally, I would suggest jumping right from completing the square to trig sub.
I don't think that u-sub is really necessary.
I guess it makes things a little easier to read though :o

Answer 7

so where do I go after u sub because I still have the problem of the square root on the bottom

Answer 8

what is a trig sub?

Answer 9

\[\Large\bf\sf \int\limits \frac{1}{\sqrt{u^2-9}}\;du\]Can you see the math equation I posted above? Or is it just showing up as ugly code?

Answer 10

ugly code idk why

Answer 11

Grrr.. :(

Answer 12

You haven't learned about Trigonometric Substitutions yet? :o
Hmm that's strange.

Answer 13

is that the only way I can solve it from here?

Answer 14

Ya I don't know of any other method off hand. Hmm

Answer 15

I can show you, it might be a little tricky if you've never seen it before though.

Answer 16

@ zepdrix, I challenge you to demo and explain your method, in as straightforward a fashion as possible.

Answer 17

I can't do it in this awful text :(
Imma draw up some instructions for you in photoshop, gimme few mins.

Answer 18

By the way, zepdrix' suggestion, "Personally, I would suggest jumping right from completing the square to trig sub.
I don't think that u-sub is really necessary." is entirely appropriate. more power to him. But he still has the burden of showing the rest of us how to do things his way (and to get it right).

Answer 19

Criticism, constructive or not, warrants challenges to demonstrate a better way. :)

Answer 20

ok

Answer 21

alright that make sense

Answer 22

So we can plug that into our denominator, and it will clean up like that.
But we still need to deal with the differential dx.
We need to get it in terms of u.

Do you remember the derivative of secant?

Answer 23

Nice work zepdrix. Regarding that 9 under the radical sign:

why not rewrite u^2 - 9 as 9( (u/3)^2 - 1)? Please experiment with that and see if it makes sense.

Answer 24

Factor the 9 out ahead of time?
Ya that's a good approach also.

Answer 25

dy/dx(sec)=secxtanx

Answer 26

Obviously the square root of that would be 3 Sqrt( (u/3)^2 -1. Certainly you can still use a trig substitution.

Answer 27

Ok good.

So our u = 3secθ gives us

du = 3secθ tanθ dθ

Answer 28

ok so does that mean we have now the integral 3sec

Answer 29

sorry how do you get the equation button to work?

Answer 30

I think we lose the 3 also right recon? Check my math real quick.

Answer 31

Both the `equation` and `draw` utilities are broken right now.
They should be fixed sometime this week.

Answer 32

yeah that was what I was trying to right just not simplified

Answer 33

write*

Answer 34

Remember this integral? It's a doozy if you try to work it out by hand.
Good one to memorize.

Answer 35

i have not learned the integral of solely secant theta yet

Answer 36

Ok well we don't have to time to go through the steps right now.
It's a bit of a nasty one XD heh.

Anyway the integral of secθ gives us:

ln| secθ + tanθ | + c

Answer 37

This is where things get a little tricky.
We have to go back to good ole triangle math in order to write our secθ and tanθ back in terms of u.

Answer 38

I think I might just ask my teacher tomorrow thank you for all of your help though

Answer 39

Ok np :O