wana help poor newbie at integration? (1 Viewer)

[Newbie]

integrate

1/[(x^2+a^2)^n] dx

 

[OLDMAN]

Try substitution : atan@=x. Unfortunately you may not be able to get away from a reduction formula.

 

[MR]

the integration question

that is a pretty hard integration, i had a crack at it, i dont know if its correct. i let x=atanQ as suggested, my answer was:
(note I_n will refer to the integral of I for n, similiarly for I_n-2)

I_n = {a^(1-2n)tanQ(cosQ)^2n + [2n.a^(1-2n)-1] I_n-2

as i said im not sure if its right, if u have the answer, and this is it i will send u my working.

 

[ezzy85]

i did this one today:

use by parts:

u = (x² + a²)^-n
u`= -2nx/(x² + a²)ⁿ⁺¹

v`=1
v=x

go on from there.

the answers I_n = 1/(a²(2n-2)) [x/(x² + a²)^n-1) + (2n-3)I_n-1]

 

[Affinity]

substituting x=a*tan(t) and and working with the reduction formula with cos looks easier.