The integration of x^4-1 is simple. You can integrate it piece by piece, so intx^4dx = x^(4+1)/(4+1) = x^5/5 + C and int-1dx = -x + C. We will drop
C as it is unneeded for finding a.
So what we have is x^5/5-x where the upper-bound is a and the lower-bound is -a. If we plug in a and -a into our function as per with every definite integral, we will have:
a^5/5-a-(-a^5/5+a) = 2a^5/5-2a.
Realize that our goal is to have that function equate to zero, so rewriting it, we will get:
2a^5/5-2a = 0.
So, adding 2a to both sides, we get 2a^5/5 = 2a. Then we multiply by 5 and we get 2a^5 = 10a. Then dividing by 2, we get a^5 = 5a then continuing we finally get a^4 = 5.
Then we fourth-root both sides to get a = +-root4(5).
