Integrating v^2/v^2+4 - Exploring Arctan & u Substitution

In summary, to integrate \frac{v^2}{v^2 + 4}, you can rewrite it as 1 - \frac{4}{v^2 + 4} and integrate using the formula for \int \frac{dx}{x^2+a^2}. Another possible method is to substitute u = v^2 + 4 and use integration by parts, but this may result in a more complicated solution. Ultimately, both methods will lead to the same answer of x - 2 tan^{-1}\left(\frac{x}{2}\right).
  • #1
fiziksfun
78
0

Homework Statement



how do i integrate

v^2 / v^2 + 4

Homework Equations



i understand this has something to do with arctan

but if i use u substitution to let v=(u/2) so (on the bottom) it becomes (1/4)(1+(v/2)^2)

there's still a v^2 on the top which the u substitution does not get rid of. Help :[


The Attempt at a Solution

 
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  • #2
this is a bit of a gues but why don't you write it as

[itex]\int \frac{v}{2} \frac{2v}{v^2+4} dv[/itex]

then do it by parts but for the 2nd term you can substitute [itex]u=v^2+4[/itex] giving [itex]\frac{du}{u}[/itex].

you'll get logs not arctans which i reckon is true because the your integral isn't of the form [itex]\int \frac{dx}{x^2+a^2}[/itex]
 
  • #3
Use polynomial division to rewrite [tex] \frac{v^2}{v^2 + 4} [/tex] as [tex] 1 - \frac{4}{v^2 + 4}[/tex]. Can you integrate it now?
 
  • #4
...or that lol!

would't my method work as well?
 
  • #5
latentcorpse said:
...or that lol!

would't my method work as well?

Doing it your way, you'd have to evaluate [tex] \int ln(x^2+4) [/tex] which evaluates to

[tex] 4 tan^{-1}\left(\frac{x}{2}\right) + x \left(ln(x^2+4) - 2\right) [/tex]

My way seems to be easier.

Let me summarize what happens your way, I switched v's to x's.

[tex] \int \frac{x^2}{x^2+4} dx = \int \frac{x}{2} \frac{2x}{x^2+4} dx = \frac{x}{2} ln(x^2+4) - \frac{1}{2} \int ln(x^2+4) dx = \frac{x}{2} ln(x^2+4) - \frac{1}{2} \left(4 tan^{-1}\left(\frac{x}{2}\right) + x \left(ln(x^2+4) - 2\right)\right) [/tex]

Now obviously [tex] \frac{x}{2} ln(x^2+4) [/tex] will cancel and you will be left with

[tex] x - 2 tan^{-1}\left(\frac{x}{2}\right) [/tex]

This is the exact same answer you get if you do it my way:

[tex] \int \frac{x^2}{x^2+4} dx = \int 1 - \frac{4}{x^2+4} dx = x - 4\left(\frac{1}{2} tan^{-1}\left(\frac{x}{2}\right)\right) = x - 2 tan^{-1}\left(\frac{x}{2}\right)[/tex]
 
Last edited:

Related to Integrating v^2/v^2+4 - Exploring Arctan & u Substitution

1. What is the purpose of exploring arctan and u substitution?

The purpose of exploring arctan and u substitution is to solve integrals that involve expressions of the form v^2/v^2+4. Using arctan and u substitution, we can simplify these integrals and make them easier to solve.

2. How does arctan substitution work?

Arctan substitution involves substituting a variable u for an expression that involves arctan. This allows us to rewrite the integral in terms of u, making it easier to solve. The substitution rule for arctan is: ∫f(arctan(x))dx = ∫f(u)(1+x^2)du.

3. What is u substitution and how is it related to arctan substitution?

U substitution, also known as the substitution method, is a technique used to simplify integrals by substituting a variable u for a more complex expression. This is similar to arctan substitution, as both methods involve substituting a variable to simplify the integral. However, u substitution can be used for a wider range of integrals, while arctan substitution is specifically used for integrals of the form v^2/v^2+4.

4. What are the steps for integrating v^2/v^2+4 using arctan and u substitution?

The steps for integrating v^2/v^2+4 using arctan and u substitution are as follows:

  1. Identify the integral expression as v^2/v^2+4.
  2. Use the substitution rule for arctan to rewrite the integral in terms of u.
  3. Simplify the integral using u substitution.
  4. Integrate the simplified expression and substitute back in for u.
  5. Simplify the final answer, if possible.

5. Can arctan and u substitution be used for all integrals?

No, arctan and u substitution can only be used for integrals of the form v^2/v^2+4. For other types of integrals, different techniques may need to be used. It is important to identify the form of the integral before choosing a method of integration.

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