y = x^2/2 , [0,4] Find the arc length of the curve over the given interval.

To find the arc length of a curve, we follow the formula:
S = int_a^b sqrt(1+((dy)/(dx))^2)  if y=f(x) , alt=xlt=b or [a,b] .
For the given problem: y =x^2/2 on interval [0,4] , we have boundary values: a= 0 and b=4 .  
Apply Power Rule for differentiation: d/(dx) x^n = n * x^(n-1) * dx .
(dy)/(dx) = d/(dx) (x^2/2)
      = 1/2d/(dx) (x^2)
      = 1/2 * [ 2 *x^(2-1) * 1 ]
      =1/2 * [ 2x]
    = (2x)/2
      = x
Plug-n a=0 , b = 4 , and (dy)/(dx)= x on the formula S = int_a^b sqrt(1+((dy)/(dx))^2) , we get:
S = int_0^4 sqrt(1+x^2) dx
From indefinite integral table, the problem resembles the formula for integral with roots:
int sqrt(u^2+-a^2) dx=1/2usqrt(a^2+-u^2)+-1/2a^2ln|u+sqrt(u^2+-a^2)| .
Take note we have "+ " sign inside the radical part then we follow formula as:
int sqrt(u^2+a^2)dx=1/2*usqrt(a^2+u^2)+1/2*a^2ln|u+sqrt(u^2+a^2)| .
Applying the formula, we get
S = int_0^4 sqrt(1+(x)^2)
=[1/2*xsqrt(1^2+x^2)+1/2*1^2ln|x+sqrt(x^2+1^2)|]|_0^4
=[1/2*xsqrt(1+x^2)+1/2*ln|x+sqrt(x^2+1)|]|_0^4
=[(xsqrt(1+x^2))/2+(ln|x+sqrt(x^2+1)|)/2]|_0^4
Apply definite integration formula: F(x)|_a^b= F(b)-F(a) .
S =[(4sqrt(1+4^2))/2+(ln|4+sqrt(4^2+1)|)/2]-[(0sqrt(1+0^2))/2+(ln|0+sqrt(0^2+1)|)/2]
=[(4sqrt(1+16))/2+(ln|4+sqrt(16+1)|)/2]-[(0sqrt(1+0))/2+(ln|0+sqrt(0+1)|)/2]
=[(4sqrt(17))/2+(ln|4+sqrt(17)|)/2]-[0/2+(ln|0+sqrt(1)|)/2]
=[ 2sqrt(17)+(ln|4+sqrt(17)|)/2]-[0+(ln|1|)/2]
=[ 2sqrt(17)+(ln|4+sqrt(17)|)/2]-[0+0/2]
=[ 2sqrt(17)+(ln|4+sqrt(17)|)/2]-[0]
=2sqrt(17)+(ln|4+sqrt(17)|)/2  or 9.29 (approximated value)