Recall indefinite integral follows `int f(x) dx = F(x)+C`

where:

`f(x)` as the integrand

`F(x)` as the antiderivative of f(x)

`C` as the constant of integration.

From the table of integrals, we follow the formula:

`sqrt(x^2+-a^2) dx = 1/2xsqrt(x^2+-a^2)+-1/2a^2ln|x+sqrt(x^2+-a^2)|`

From the given problem `int_0^3 sqrt(x^2+16) dx` , we have a addition sign (+) in between terms inside the square root sign. Then, we follow the formula:

`int sqrt(x^2+a^2) dx = 1/2xsqrt(x^2+a^2)+1/2a^2ln|x+sqrt(x^2+a^2)|`

Take note that we can express `16 = 4^2` then the given problem becomes:`int_0^3 sqrt(x^2+4^2) dx` .

The `x^2 +4^2` resembles the `x^2 +a^2` in the formula. Then by comparison, the corresponding values are: `x=x` and `a=4.`

Plug-in `x=x` and `a=4` on the formula, we get:

`int_0^3 sqrt(x^2+16) dx ` `=[1/2xsqrt(x^2+4^2)+1/2*4^2ln|x+sqrt(x^2+4^2)| ]|_0^3`

`=[1/2xsqrt(x^2+16)+1/2*16ln|x+sqrt(x^2+16)|]|_0^3`

`=[1/2xsqrt(x^2+16)+8ln|x+sqrt(x^2+16)|]|_0^3`

Apply definite integral formula: `F(x)|_a^b = F(b) - F(a)` .

`[1/2xsqrt(x^2+16)+8ln|x+sqrt(x^2+16)|]|_0^3`

`=[1/2*3sqrt(3^2+16)+8ln|3+sqrt(3^2+16)|]-[1/2*0sqrt(0^2+16)+8ln|0+sqrt(0^2+16)|]`

`=[3/2sqrt(9+16)+8ln|3+sqrt(9+16)|]-[0*sqrt(0+16)+8ln|0+sqrt(0+16)|]`

`=[3/2*5+8ln|3+5|]-[0*4+8ln|0+4|]`

`=[15/2+8ln|8|]-[0+8ln|4|]`

`=15/2+8ln|8| -0-8ln|4|`

`=15/2+8ln|8| - 8ln|4|`

`=15/2+8(ln|8| - ln|4|)`

Apply natural logarithm property: `ln(x)- ln(y) = ln(x/y)` .

`=15/2+8ln|8/4|`

`=15/2+8ln|2|`

Apply natural logarithm property: ` n*ln(x) = ln(x^n)` .

`=15/2+ln|2^8|`

=`15/2+ln|256|` or `13.05` ( approximated value)

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