Can someone help me caluclate the arc length of x^2 from -4 to 4

y = x^2
dy/dx = 2x

sqrt(1 + (2x)^2) * dx =>
sqrt(1 + 4x^2) * dx

x = (1/2) * tan(t)
dx = (1/2) * sec(t)^2 * dt

sqrt(1 + 4 * (1/4) * tan(t)^2) * (1/2) * sec(t)^2 * dt =>
sqrt(1 + tan(t)^2) * (1/2) * sec(t)^2 * dt =>
(1/2) * sqrt(sec(t)^2) * sec(t)^2 * dt =>
(1/2) * sec(t) * sec(t)^2 * dt =>
(1/2) * sec(t)^3 * dt

Now, let's find the integral of sec(t)^3 * dt

int(sec(t)^3 * dt) =>
int(sec(t) * sec(t)^2 * dt)

u = sec(t)
du = sec(t) * tan(t) * dt
dv = sec(t)^2 * dt
v = tan(t)

int(sec(t)^3 * dt) = sec(t) * tan(t) - int(tan(t)^2 * sec(t) * dt)
int(sec(t)^3 * dt) = sec(t) * tan(t) - int((sec(t)^2 - 1) * sec(t) * dt)
int(sec(t)^3 * dt) = sec(t) * tan(t) - int(sec(t)^3 * dt) + int(sec(t) * dt)
2 * int(sec(t)^3 * dt) = sec(t) * tan(t) + int(sec(t) * dt)
2 * int(sec(t)^3 * dt) = sec(t) * tan(t) + ln|sec(t) + tan(t)| + C
int(sec(t)^3 * dt) = (1/2) * (sec(t) * tan(t) + ln|sec(t) + tan(t)| + C

So...

(1/2) * int(sec(t)^3 * dt) = (1/2) * (1/2) * (sec(t) * tan(t) + ln|sec(t) + tan(t)| + C
(1/2) * int(sec(t)^3 * dt) = (1/4) * (sec(t) * tan(t) + ln|sec(t) + tan(t)| + C

x = (1/2) * tan(t)
2x = tan(t)

x = (1/2) * tan(t)
x^2 = (1/4) * tan(t)^2
4x^2 = tan(t)^2
4x^2 = sec(t)^2 - 1
4x^2 - 1 = sec(t)^2
sec(t) = sqrt(4x^2 - 1)

(1/4) * (sqrt(4x^2 - 1) * 2x + ln|sqrt(4x^2 - 1) + 2x|) + C

From x = -4 to x = 4

(1/4) * (sqrt(4 * 4^2 - 1) * 2 * 4 + ln|sqrt(4 * 4^2 - 1) + 2 * 4|) - (1/4) * (sqrt(4 * (-4)^2 - 1) * 2 * (-4) + ln|sqrt(4 * (-4)^2 - 1) + 2 * (-4)|)

(1/4) * (sqrt(64 - 1) * 8 + ln|sqrt(64 - 1) + 8| - sqrt(64 - 1) * (-8) - ln|sqrt(64 - 1) - 8|) =>
(1/4) * (sqrt(63) * 8 + ln|sqrt(63) + 8| - sqrt(63) * (-8) - ln|sqrt(63) - 8|) =>
(1/4) * (16 * sqrt(63) + ln|(sqrt(63) + 8) / (sqrt(63) - 8)|) =>
(1/4) * (16 * 3 * sqrt(7) + ln|(63 + 16 * sqrt(63) + 64) / (63 - 64)|) =>
(1/4) * (48 * sqrt(7) + ln|(127 + 16 * 3 * sqrt(7)) / (-1)|) =>
(1/4) * (48 * sqrt(7) + ln(127 + 48 * sqrt(7)))