# Consider the functions: F(x)= definite integral [1,x] of f(t)dt f(t)= definite integral [1,t^2] of (sqrt(5+u^4)/u) du Find F''(1). Use the Fundamental Theorem of Calculus

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You need to evaluate f(t), hence, you need to solve the definite integral `int_1^(t^2)(sqrt(5+u^4))/u du` such that:

`u^4 =v => 4u^3du = dv => du = (dv)/(4u^3)`

`int (sqrt(5+u^4))/u du = int (sqrt(5+v))/u(dv)/(4u^3) `

`int (sqrt(5+v))/(4u^4) dv = (1/4)int (sqrt(5+v))/v(dv)`

You should consider the next substitution such that:

`sqrt(5+v) = s => s^2 = 5+v => v = s^2-5`

`1/(2sqrt(5+v))dv = ds => dv = 2sqrt(5+v)ds => dv = 2s ds`

`(1/4)int (sqrt(5+v))/v(dv) = (1/2)int s^2/(s^2-5)ds`

You need to subtract and add 5 to numerator such that:

`int (s^2-5+5)/(s^2-5)ds `

Using the property of linearity yields:

`int (s^2-5+5)/(s^2-5)ds = int (s^2-5)/(s^2-5)ds + int (5)/(s^2-5)ds `

`int (s^2-5+5)/(s^2-5)ds = int ds + 5 ln|(s-5)/(s+5)| + c`

`(1/2)int s^2/(s^2-5)ds = s/2 + (5/2)ln|(s-5)/(s+5)| + c`

Substituting back `sqrt(5+v)` for s yields:

`(1/4)int (sqrt(5+v))/v(dv) = (sqrt(5+v))/2 + (5/2)ln|(sqrt(5+v)-5)/(sqrt(5+v)+5)| + c`

Substituting back `u^4` for v yields:

`int (sqrt(5+u^4))/u du = (sqrt(5+u^4))/2 + (5/2)ln|(sqrt(5+u^4)-5)/(sqrt(5+u^4)+5)| + c `

Using the fundamental theorem of calculus yields:

`int_1^(t^2) (sqrt(5+u^4))/u du = ((sqrt(5+u^4))/2 + (5/2)ln|(sqrt(5+u^4)-5)/(sqrt(5+u^4)+5)|)|_1^(t^2)`

`int_1^(t^2) (sqrt(5+u^4))/u du = ((sqrt(5+t^8))/2 + (5/2)ln|(sqrt(5+t^8)-5)/(sqrt(5+t^8)+5)| - sqrt(6))/2- (5/2)ln|(sqrt(6)-5)/(sqrt(6)+5))`

Hence, evaluating f(t) yields`f(t) = ((sqrt(5+t^8))/2 + (5/2)ln|(sqrt(5+t^8)-5)/(sqrt(5+t^8)+5)| - sqrt(6))/2 - (5/2)ln|(sqrt(6)-5)/(sqrt(6)+5)).`

You need to remember that `F'(x) = f(x)` and `F''(x) = f'(x), ` hence, you need to evaluate f'(1) to find F''(1) such that:

`f'(t) = (((sqrt(5+t^8))/2)' + (5/2)(ln|(sqrt(5+t^8)-5)| - ln|(sqrt(5+t^8)+5)|)'`

`f'(t) = 2t^7/(sqrt(5+t^8)) + 10(1/(sqrt(5+t^8)-5))(t^7)/(sqrt(5+t^8))- 10(1/(sqrt(5+t^8)+5))(t^7)/(sqrt(5+t^8))`

You need to evaluate f'(1) such that:

`f'(1) = 2/(sqrt 6) + 10(1/(sqrt(6)-5))(1/(sqrt(6))) - 10(1/(sqrt(6)+5))(1/(sqrt(6)))`

`f'(1) = 2/(sqrt 6) + 10/(sqrt 6)(1/(sqrt 6 - 5) - 1/(sqrt 6 + 5))`

`f'(1) = 2/(sqrt 6) + 10/(sqrt 6)(sqrt6 + 5 - sqrt6 + 5)/(6-25)`

`f'(1) = 2/(sqrt 6)(1- 50/19) => f'(1) = 2/(sqrt 6)(-31/19)`

`f'(1) = -62/(19sqrt6) => f'(1) = -62sqrt6/(19*6)`

`f'(1) = -31sqrt6/(19*3) => f'(1) = -31sqrt6/57 = F''(1)`

**Hence, evaluating `F''(1)` under the given conditions yields `F''(1) = -31sqrt6/57.` **