Recurrence relation for an integral equation

[aperception]

Hi guys, would appreciate any help with this problem. It's from a graduate school entrance examination which I am practicing.

problem statement

When n is a natural number, the indefinite integral I_n is defined as

I_n=\int\frac{1}{(x^2+a^2)^n}dx

Here, a is a constant real number, and not equal to zero. Answer the following questions.

1. Express I_{n+1} as a recurrence equation with I_n.
2. Derive I_1 and I_2

where I'm at
I can solve part 2 relatively easily by using the trigonometric substitution x=a \tan\theta which gives the result (feel free to ask for working if you don't believe me):

I_n=\frac{1}{a^{2n-1}}\int(\cos\theta)^{2n-2}d\theta

or:

I_{n+1}=\frac{1}{a^{2n+1}}\int(\cos\theta)^{2n} d\theta

these are fairly easy to solve for fixed values of n using basic integration and backwards substitutions such as \tan\theta=(x/a) etc, giving:

I_1=\arctan(x/a)/a

I_2=\arctan(x/a)/(2a^3)+\frac{x}{2a^2(x^2+a^2)}

but I don't know how to generalize to the final recurrence equation.

attempted solution
I tried to solve the original equation using integration by parts by writing:

I_{n+1}=\int\frac{1}{(x^2+a^2)^n}\frac{1}{(x^2+a^2)}dx

then setting:
u=\frac{1}{(x^2+a^2)}, v'=\frac{1}{(x^2+a^2)^n}


u'=\frac{-2x}{(x^2+a^2)^2}, v=I_n

this gives:
I_{n+1}=\frac{I_n}{(x^2+a^2)}+\int\frac{I_n 2x}{(x^2+a^2)^2}dx

but this doesn't agree with the calculations from above with fixed n, so it must be wrong. (I presume v=I_n is invalid for some reason?).

Unfortunately I don't really have any ideas on where to go from here...?

 

[SammyS]

Try the following: (I haven't checked it out myself, so it may be bogus.)

\displaystyle I_n=\int\frac{1}{(x^2+a^2)^n}dx

\displaystyle =\int\frac{x^2+a^2}{(x^2+a^2)^{(n+1)}}dx

\displaystyle =a^2I_{n+1}+\int (x)\,\frac{x}{(x^2+a^2)^{(n+1)}}dx

Now try integration by parts: u = x, \displaystyle dv=\frac{x}{(x^2+a^2)^{(n+1)}}dx​

 

[aperception]

That is very clever, it worked great thank you!