Find recursive forumula for integrals I have to find recursve formulas for solving the following two integrals. The assignment tells one to find an Expression that leads from the calculation of $\dfrac{I_{2n}}{I_{2n+1}}$ to the calculation of $\dfrac{I_{2n-2}}{I_{2n-1}}$
a) $$ I_{2n} = \int_a^b \frac{1}{(1+x^2)^n} dx   $$
b) $$ I_{2n+1} = \int_a^b \frac{1}{\sqrt{1+x^2}^{2n+1}} dx   $$
This Problem sits in my mind for a few days now and I am really stuck with it. The idea I have come up with for a) is to use Integration by parts:
Let u' be $\frac{1}{1+x^2} $ and v be $\frac{1}{(1+x^2)^{n-1}}$;
u then is $tan^{-1}x$ and I could write:
$$ I_{2n} = \int_a^b \frac{1}{(1+x^2)^n} dx  = \int_a^b  \frac{1}{1+x^2} \frac{1}{(1+x^2)^{n-1}} dx$$
$$=\tan^{-1}x\cdot\frac{1}{(1+x^2)^{n-1}} -  \int_a^b \tan^{-1}x\cdot \frac{d}{dx} \frac{1}{(1+x^2)^{n-1}} dx $$
I could repeat using Integration by parts on the right side but it did not lead me to anything useful...
Do you have any Idea on how to solve this?
Thank you for your help in advance!
FunkyPeanut

I could show one way to derive the recursive formulas given in one of the comments:
Write
$$I_{2n}=\int\frac{1+x^2-x^2}{(1+x^2)^n}dx= I_{2n-2}-\int\frac{x^2}{(1+x^2)^n}dx
$$
We have to compute the last integral. Using integration by parts (with $u'=x/(1+x^2)^n$) we have
$$
\int\frac{x^2}{(1+x^2)^n}dx= \frac{x}{(2-2n)(1+x^2)^{n-1}}-\frac{1}{2-2n}I_{2n-2}
$$
Hence,
$$
I_{2n}= I_{2n-2}+\frac{x}{(2n-2)(1+x^2)^n}-\frac{1}{2n-2}I_{2n-2}= \frac{2n-1}{2n-2}I_{2n-2}+\frac{x}{(2n-2)(1+x^2)^n}
$$
For the odd indices, we can use the "same trick" to get
$$
I_{2n+1}=\int\frac{1+x^2-x^2}{\sqrt{(1+x^2)^{2n+1}}}dx= I_{2n-1}-\int\frac{x^2}{\sqrt{(1+x^2)^{2n+1}}}dx
$$
Again, integration by parts yields (with $u'=x/\sqrt{(1+x^2)^{2n+1}}$)
$$
 \int\frac{x^2}{\sqrt{(1+x^2)^{2n+1}}}dx= \frac{x}{(1-2n)\sqrt{(1+x^2)^{2n-1}}}-\frac{1}{1-2n}I_{2n-1}
$$
Therefore, after some reductions,
$$
I_{2n+1}= \frac{2n}{2n-1}I_{2n-1}+\frac{x}{(2n-1)\sqrt{(1+x^2)^{2n-1}}}.$$
