Help finding the indefinite integral

In summary, you were able to use the integration formula to solve for this du, but you needed to find the u.
  • #1
Chandasouk
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
 
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  • #3
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
Don't forget to use the property of...

[tex]\sqrt{a b} = \sqrt{a}\sqrt{b}[/tex]
 
  • #5
How about u = sqrt(3)x, which makes du = sqrt(3)dx?
 
  • #6
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
Chan, Mark said to use...

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

You're using there...

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

Note the difference.
 
  • #8
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
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
[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.
 

Related to Help finding the indefinite integral

1. What is an indefinite integral?

An indefinite integral is a mathematical operation that is used to find the most general antiderivative of a given function. It is also known as the reverse process of differentiation.

2. How do I find the indefinite integral of a function?

To find the indefinite integral of a function, you can use a variety of techniques such as the power rule, substitution method, integration by parts, and partial fractions. It is important to remember the constant of integration when evaluating the indefinite integral.

3. Why is finding the indefinite integral important?

Finding the indefinite integral is important in many areas of science and engineering, as it allows us to calculate the total change or the accumulation of a quantity over a given interval. It is also a fundamental concept in calculus and is used to solve various mathematical problems.

4. Can a function have multiple indefinite integrals?

Yes, a function can have multiple indefinite integrals, as the constant of integration can take on different values. This means that two indefinite integrals of the same function may differ by a constant.

5. Is there a shortcut for finding the indefinite integral?

There is no single shortcut for finding the indefinite integral of a function. However, with practice and familiarity with integration techniques, you can develop a better understanding of which method may be most efficient for a given function and save time in the integration process.

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