Help finding the indefinite integral

  • Thread starter Chandasouk
  • Start date
  • #1
165
0
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
 

Answers and Replies

  • #3
165
0
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)
 
  • #4
Char. Limit
Gold Member
1,208
14
Don't forget to use the property of...

[tex]\sqrt{a b} = \sqrt{a}\sqrt{b}[/tex]
 
  • #5
35,231
7,051
How about u = sqrt(3)x, which makes du = sqrt(3)dx?
 
  • #6
165
0
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?
 
  • #7
Char. Limit
Gold Member
1,208
14
Chan, Mark said to use...

[tex]u = \sqrt{3} x[/tex]

You're using there...

[tex]u = \sqrt{3x}[/tex]

Note the difference.
 
  • #8
165
0
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)
 
  • #9
35,231
7,051
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
165
0
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
35,231
7,051
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
165
0
Are there any tricks to find out the U's?
 
  • #13
35,231
7,051
With some practice it's pretty straightforward. You were trying to match your integral
[tex]\int \frac{dx}{4 + 3x^2}[/tex]
to an entry in a table of integrals like this:
[tex]\int \frac{dx}{a^2 + u^2}[/tex]

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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