Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 46

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)