Question

integration of D^n(z^2+1)^2n​

Answer 1

Answer:

dn

dxn

(x−1)n(x+1)n=

n

∑

k=0 (

n

k

)(

dk

dxk

(x−1)n)(

dn−k

dxn−k

(x+1)n)

Now, to evaluate Rn(1), just observe that for 0≤k≤n−1 the term (

dk

dxk

(x−1)k) still contains an x−1. Thus

Rn(1)=(

dn

dxn

(x−1)n)(x+1)n|x=1=n!2n

P.S. The above also Yields:

dn

dxn

(x−1)n(x+1)n=

n

∑

k=0 (

n

k

)(

n!

(n−k)!

(x−1)n−k)