How to calculate this? #int_0^a(a-x)^(n-1)/(a+x)^(n+1)dx# ;#a>0;ninNN#*.

1 Answer
Cesareo R.
May 16, 2017

#1/(2an)#

Explanation:

#d/(dx)((a-x)/(a+x))^n = -2an((a-x)/(a+x))^n/(a^2-x^2) = -2an (a-x)^(n-1)/(a+x)^(n+1)# so

#(a-x)^(n-1)/(a+x)^(n+1)=-1/(2an)d/dx((a-x)/(a+x))^n# and then

#int_0^a(a-x)^(n-1)/(a+x)^(n+1)dx=-1/(2an)[((a-x)/(a+x))^n]_0^a = 1/(2an)#

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