Question

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

Answer 1

`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)`