Help finding the indefinite integral

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Homework Help Overview

The discussion revolves around finding the indefinite integral of the function 1/(4+3x^2), with participants exploring substitution methods related to integration formulas.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various substitutions for u, including u = 3x and u = sqrt(3)x, and question the implications of their choices on the differential du. There is also exploration of how to simplify expressions involving derivatives.

Discussion Status

Several participants have offered different approaches to the substitution, with some clarifying the differences between their proposed u values. There is an ongoing exploration of how to correctly apply the integration formula and check the results.

Contextual Notes

Participants are navigating through the constraints of matching the integral to known forms and are discussing the challenges of identifying appropriate substitutions. There is a mention of needing to practice to become more proficient in recognizing suitable u values.

Chandasouk
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1/(4+3x^2) ?

I'm trying to use this integration formula to solve for this

du
----------
a^2+u^2

I know my a would be 2, but what would my u be? I tried u = 3x and du=3dx and du/3=dx

but (3x)^2 would be 9x^2
 
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You want u2 = 3x2, so what should u be?
 
sqrt(3x^2)

but then du would be 1/2(3x^2)^(-1/2)*6x dx

which simplifies to

6x dx
-------------
2*sqrt(3x^2)
 
Don't forget to use the property of...

\sqrt{a b} = \sqrt{a}\sqrt{b}
 
How about u = sqrt(3)x, which makes du = sqrt(3)dx?
 
Mark

if U is sqrt(3x) which can be written as 3x^1/2

when I find the du, it becomes

1/2*(3x)^(-1/2)*3 dx

which is written as

3 dx
------
2*sqrt(3x)

I'm trying to just get du to equal dx so I can use the formula. Should I just multiply the denominator over and divide by 3?
 
Chan, Mark said to use...

u = \sqrt{3} x

You're using there...

u = \sqrt{3x}

Note the difference.
 
ooooh, sorry.

When I get du = sqrt(3)dx, I just got dx by itself by dividing.

It was plug and chug from there

1
------- arctan(sqrt(3)*x) / 2 + C
2*sqrt(3)
 
Chandasouk said:
ooooh, sorry.

When I get du = sqrt(3)dx, I just got dx by itself by dividing.

It was plug and chug from there

1
------- arctan(sqrt(3)*x) / 2 + C
2*sqrt(3)

I didn't check, but it looks like you might have an extra 2. You can check by differenting your answer.
 
  • #10
Hmm, i think i followed the template

1/a arctan (u/a) + C

1 1
-- * ----- arctan(sqrt(3)*x/2) + C
2 sqrt(3)
 
  • #11
Chandasouk said:
Hmm, i think i followed the template

1/a arctan (u/a) + C

1 1
-- * ----- arctan(sqrt(3)*x/2) + C
2 sqrt(3)
That's better. What you had before was different.
1
------- arctan(sqrt(3)*x) / 2 + C
2*sqrt(3) .
 
  • #12
Are there any tricks to find out the U's?
 
  • #13
With some practice it's pretty straightforward. You were trying to match your integral
\int \frac{dx}{4 + 3x^2}
to an entry in a table of integrals like this:
\int \frac{dx}{a^2 + u^2}

You were able to figure out that a = 2, but stumbled with you substitution for u. In this case you want u2 = 3x2, so u = sqrt(3)*x. I can't think of any tips or tricks that would be better than getting practice.
 

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