How can I integrate v² / (v² + 4) using arctan and 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)