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

[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

 

[latentcorpse]

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 $\int \frac{dx}{x^2+a^2}$

 

[NoMoreExams]

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

 

[latentcorpse]

...or that lol!

would't my method work as well?

 

[NoMoreExams]

...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:

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