# `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.

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