To be able to solve for definite integral, we follow the first fundamental theorem of calculus: `int_a^b f(x) dx = F(x) +C`

such that f is continuous and F is the antiderivative of f in a closed interval `[a,b]` .

The `[a.b]` is the boundary limits of the integral such as lower bound=a and upper bound = b.

For the given problem:` int_3^(6) 1/(25+(x-3)^2)dx` ,

it resembles the basic integration formula:

`int (du)/(a^2+u^2) =(1/a)arctan(u/a)+C` .

By comparison: `(du)/(a^2+u^2) vs(1/(25+(x-3)^2))dx` , we may apply

u-substitution by letting:

`u^2=(x-3)^2` then `u = x-3`

where `a^2=25 or 5^2` then `a=5`

Derivative of u will be `du = 1 dx` or `du = dx` .

`int_3^(6) 1/(25+(x-3)^2)dx =int_3^(6) 1/(25+(u)^2)du`

Applying the formula:

`int_3^(6) 1/(25+(u)^2)du =(1/5)arctan(u/5)|_3^6`

Plug-in `u = x-3` to express the indefinite integral in terms of x:

`(1/5)arctan(u/5)|_3^6 =(1/5)arctan((x-3)/5)|_3^6`

Recall `F(x)|_a^b = F(b) - F(a)` then:

`(1/5)arctan((x-3)/5)|_3^6 = F(6)-F(3)`

` = (1/5)arctan((6-3)/5) -(1/5)arctan((3-3)/5)`

`= (1/5)arctan(3/5) -(1/5)arctan(0/5)`

` =(1/5)arctan(3/5) -0 `

`=(1/5)arctan(3/5) ` as the** Final Answer**.

Note: `arctan(0/5) = arctan(0)= 0 `

since `tan(theta) = 0` when `theta=0`

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