# Recursive integral using integration by pars

1. Jan 31, 2012

### oferon

First excuse my bad english on math subjects. I'm working on it.

How can I integrate by parts:
$$I_{m}=\int\frac{1}{(x^2+a^2)^m}\,dx$$

I need to find a recursive form,
But I can't find the right g' and f to get this done...

I've tried
$$g'=1 \quad\,\quad\ f=\frac{1}{(x^2+a^2)^m}$$
As well as $$g' = \frac{1}{(x^2+a^2)}\quad\ →g=arctan(x/a)\ , \quad\ f=\frac{1}{(x^2+a^2)^{m-1}}$$

But on the next integral of g*f ' , I can't find any way to simplify it to Im-1 or another integration by parts that will lead somewhere.

Which g' and f should I pick for this integral then? Thanks in advance..

Last edited: Jan 31, 2012
2. Jan 31, 2012

### tiny-tim

welcome to pf!

hi oferon! welcome to pf!

(btw, your english is fine … on this post at least )

(i haven't tried it myself , but …)

my guess is that the trick is to write it (x2 + a2)/(x2 + a2)m+1,

and then use x as f or g

3. Jan 31, 2012

### oferon

Which "x" do you reffer to saying "and then use x as f or g" ?

I've tried the trick you suggested using:

$$g'=x^2+a^2 \quad\, \quad\ f=\frac{1}{(x^2+a^2)^{m+1}}$$

but all I get is:

$$\frac{arctan(\frac{x}{a})}{(x^2+a^2)^{m+1}} - ∫arctan(\frac{x}{a})*\frac{-2x(m+1)}{(x^2+a^2)^{m+2}}$$

And now I'm stuck all over again..

4. Jan 31, 2012

### tiny-tim

hi oferon!

hint: x2/(x2 + a2) = x(x/(x2 + a2))