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

• fiziksfun
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).
fiziksfun

## 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

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

$\int \frac{v}{2} \frac{2v}{v^2+4} dv$

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

you'll get logs not arctans which i reckon is true because the your integral isn't of the form $\int \frac{dx}{x^2+a^2}$

Use polynomial division to rewrite $$\frac{v^2}{v^2 + 4}$$ as $$1 - \frac{4}{v^2 + 4}$$. Can you integrate it now?

...or that lol!

would't my method work as well?

latentcorpse said:
...or that lol!

would't my method work as well?

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

$$4 tan^{-1}\left(\frac{x}{2}\right) + x \left(ln(x^2+4) - 2\right)$$

My way seems to be easier.

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

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

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

$$x - 2 tan^{-1}\left(\frac{x}{2}\right)$$

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

$$\int \frac{x^2}{x^2+4} dx = \int 1 - \frac{4}{x^2+4} dx = x - 4\left(\frac{1}{2} tan^{-1}\left(\frac{x}{2}\right)\right) = x - 2 tan^{-1}\left(\frac{x}{2}\right)$$

Last edited:

## 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.

• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
10
Views
1K
• Calculus and Beyond Homework Help
Replies
21
Views
1K
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
390
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
12
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
852
• Calculus and Beyond Homework Help
Replies
44
Views
4K