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Can anyone do this (calc2)

T

Trezza

Senior member
Do you know how to integrate this problem and if so please explain it to me.

~ 1/ (9x^2+12x-5)^1/2

*Note ~ means integral sign
 
If that's the integral sign, then yikes. I hope that's not calc 2, becasue I just wrote a test yesterday on that and Im in my last year of high school.
 
Originally posted by: speg
If that's the integral sign, then yikes. I hope that's not calc 2, becasue I just wrote a test yesterday on that and Im in my last year of high school.
Click to expand...

The integral sign is like a wierd capital s but straigtened i have no clue how to do it on a computer.
 
it has literally been 3 years since i touched this stuff, so caution... but

maybe factor the denominator into (3x -1)(3x+5) so that you can use the chain rule on (3x-1)^{-.5} and (3x+5)^{-.5}?

so say u = the first
dv = the second

then take derivative of u, and integrate dv (use substitution?), then original integral is just uv - ~vdu
 
Originally posted by: gopunk
it has literally been 3 years since i touched this stuff, so caution... but

maybe factor the denominator into (3x -1)(3x+5) so that you can use the chain rule on (3x-1)^{-.5} and (3x+5)^{-.5}?

so say u = the first
dv = the second

then take derivative of u, and integrate dv (use substitution?), then original integral is just uv - ~vdu
Click to expand...

you can't factor it into (3x-1)(3x+5) you have to complete teh square to make the equation something like
~1/((9x^2 +12x -12)+7)^1/2 and then do something.

SocrPlyr can you explain how you did that?
 
Originally posted by: Trezza
Originally posted by: gopunk
it has literally been 3 years since i touched this stuff, so caution... but

maybe factor the denominator into (3x -1)(3x+5) so that you can use the chain rule on (3x-1)^{-.5} and (3x+5)^{-.5}?

so say u = the first
dv = the second

then take derivative of u, and integrate dv (use substitution?), then original integral is just uv - ~vdu
Click to expand...

you can't factor it into (3x-1)(3x+5) you have to complete teh square to make the equation something like
~1/((9x^2 +12x -12)+7)^1/2 and then do something.

SocrPlyr can you explain how you did that?
Click to expand...

eh... i'm not seeing why you can't factor the polynomial inside the sqrt
 
Originally posted by: gopunk
Originally posted by: Trezza
Originally posted by: gopunk
it has literally been 3 years since i touched this stuff, so caution... but

maybe factor the denominator into (3x -1)(3x+5) so that you can use the chain rule on (3x-1)^{-.5} and (3x+5)^{-.5}?

so say u = the first
dv = the second

then take derivative of u, and integrate dv (use substitution?), then original integral is just uv - ~vdu
Click to expand...

you can't factor it into (3x-1)(3x+5) you have to complete teh square to make the equation something like
~1/((9x^2 +12x -12)+7)^1/2 and then do something.

SocrPlyr can you explain how you did that?
Click to expand...

eh... i'm not seeing why you can't factor the polynomial inside the sqrt
Click to expand...

err nevermind it appears my math skills truely are the suck
 
Originally posted by: SocrPlyr
1/3 * ln((9x^2+12x-5)^1/2 + 3x + 2)
Edit:
can't type
Click to expand...

Almost right. It is 1/3 * ln(abs((9x^2+12x-5)^1/2 + 3x + 2)) + C

The integral you are using here is
integral(du/(sqrt(u^2 - a^2)))
not
integral(du/(sqrt(u^2 + a^2)))
 
Originally posted by: Kyteland
Originally posted by: SocrPlyr
1/3 * ln((9x^2+12x-5)^1/2 + 3x + 2)
Edit:
can't type
Click to expand...

Almost right. It is 1/3 * ln(abs((9x^2+12x-5)^1/2 + 3x + 2)) + C

The integral you are using here is
integral(du/(sqrt(u^2 - a^2)))
not
integral(du/(sqrt(u^2 + a^2)))
Click to expand...

whats abs and through what process did you get this
 
abs() = absolute value function. The process I went through is I pulled you my college calc book and looked up the formula.

1) start with integral(dx/ sqrt(9x^2+12x-5))
2) do algebra to get integral(dx/sqrt((3x+2)^2-9))
3) set u=3x+2 and du = 3dx and plug in to equation to get integral(du/3sqrt(u^2-9))
4) look up formula for integral(du/sqrt(u^2-a^2)) **see above**
5) convert u back to x and get 1/3 * ln(abs((9x^2+12x-5)^1/2 + 3x + 2)) + C

final answer: 1/3 * ln(abs((9x^2+12x-5)^1/2 + 3x + 2)) + C

 
Originally posted by: Trezza
through what process did you get this
Click to expand...
I'm assuming they are looking up the answer in a table. There are plenty of books that list the answer to all common integrals like you posted. Just look up the type of equation and you are done. Or you can work it out, here is one method using a trick. Edit yep, they looked it up in a book (posted 1 minute before me).

Note: I'll use the symbol, |, as a integral sign.

Complete the square:
|(9x+12x-5)^-0.5 dx = | [(3x+2)^2 - 9]^-0.5 dx

Lets simplify this equation. Let u=3x+2. Thus you get
| [u^2 - 9]^-0.5 dx

Well du/dx = 3. Thus dx = du/3 and you get:
(1/3) | [u^2 - 9]^-0.5 du

Now for a trick (don't ask me where the trick comes from, I don't know, I just use it). Let y = [(u^2-9)^0.5+u]. Multiply and divide by y and the result is:
(1/3) | [1+u(u^2 - 9)^-0.5]/[(u^2-9)^0.5+u] du

Yes this looks like crap, but it will simplify quickly. Note that dy/du = [1+u(u^2 - 9)^-0.5]. Thus du = dy / [1+u(u^2 - 9)^-0.5]. Plug that into the equation:
(1/3) | 1/y dy

We all should be able to do that integral:
(1/3) ln(y) + C

Where C is an arbitrary constant. Plug back in the definition of y:
(1/3) ln[(u^2-9)^0.5+u] + C

Plug back in the definition of u:
(1/3) ln[({3x+2}^2-9)^0.5+3x+2] + C

Expanding the square:
(1/3) ln[(9x^2+12x-5)^0.5+3x+2] + C
 
The absolute value is optional for most uses. That just prevents the natural log from returning an imaginary number. However if you take the real part of that imaginary number it is the same thing as using the absolute value.

ln(2)=0.693
ln(abs(-2))=0.693
ln(-2)=0.693+3.142i

Notice the real part of ln(2) is the same as the real part of ln(-2).
 
Originally posted by: Kyteland
dullard (3x+2)^2-7 != 9x+12x-5

it should be (3x+2)^2-9. otherwise you are absolutely correct.
Click to expand...
Doh. I typed it too quickly. There is a reason I'm a dullard. Then copy and paste and I'm screwed all the way through.
 
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