int_0^4 1/(25-x^2) dx Evaluate the integral

Recall that first fundamental theorem of calculus indicates that int_a^b f(x) dx = F(x)|_a^b:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
"a " as the lower boundary value of x
"b " as the upper boundary value of x
To evaluate the given problem: int_0^(4)1/(25-x^2)dx , we may rewrite in a form of:
int_0^(4)1/(5^2-x^2)dx .
 The integral part resembles the integration formula for inverse of hyperbolic tangent function: int 1/(a^2-u^2) du =(1/a)arctanh(u/a)+ C.
By comparison, it shows that a^2 corresponds to 5^2 and u^2 corresponds to x^2 . Therefore, it shows that a=5 and u=x .
By following the formula, the indefinite  integral function will be:
 int_0^(4)1/(5^2-x^2)dx =(1/5)arctanh(x/5)|_0^4
To solve for the definite integral, we may apply  F(x)|_a^b= F(b)-F(a) , we get:
(1/5)arctanh(x/5)|_0^4 =(1/5)arctanh(4/5) -(1/5)arctanh(0/5)
                                =(1/5)arctanh(4/5)-(1/5)arctanh(0)
                                  =(1/5)archtan(4/5)-0
                                 = 1/5arctanh(4/5)
  The 1/5arctanh(4/5) can be simplified as  0.2197 as rounded off value.