The final answer can be seen here.
The general formula for the arc length is as follows:
D(x) = sqrt((Deltax)^2 + (Deltay)^2)
s = D(x) = sum_a^b sqrt((Deltax)^2 + (Deltay)^2/(Deltax)^2*(Deltax)^2)
= sum_a^b sqrt(1 + ((Deltay)/(Deltax))^2)Deltax
= int_a^b sqrt(1 + ((dy)/(dx))^2)dx
Thus, take the derivative and simplify.
(dy)/(dx) = 4*(1/((x/4)^2 - 1))*(2(x/4)*1/4)
= (x/((x/4)^2 - 1))*1/2
= x/(2(x/4)^2 - 2)
= x/(x^2/8 - 2)
= (8x)/(x^2 - 16)
Now, plug it in and square it:
s = int_7^8 sqrt(1 + ((8x)/(x^2 - 16))^2)dx
= int_7^8 sqrt(1 + (64x^2)/(x^2 - 16)^2)dx
= int_7^8 sqrt(((x^2 - 16)^2 + 64x^2)/(x^2 - 16)^2)dx
If we multiply this out, we should find that something like -32x^2 adds with the 64x^2 for a nice and sneaky shift into a perfect square.
(x^2 - 16)^2 = x^4 - 32x^2 + 256
(x^2 - 16)^2 + 64x^2 = x^4 + 32x^2 + 256
= (x^2 + 16)^2
Much better. Now we can get rid of that ugly square root.
= int_7^8 sqrt(((x^2+16)^2)/(x^2 - 16)^2)dx
= int_7^8 (x^2+16)/(x^2 - 16)dx
Then some manipulation to make this evaluation easier...ish.
= int_7^8 (x^2+16 - 16 + 16)/(x^2 - 16)dx
= int_7^8 (x^2 - 16 + 32)/(x^2 - 16)dx
= int_7^8 dx + int_7^8 32/((x+4)(x-4))dx
Looks like we probably have to do Partial Fraction Decomposition on this, unfortunately. Oh well.
int 32/((x+4)(x-4)) = A/(x+4) + B/(x-4)
= (A(x-4) + B(x+4))/((x+4)(x-4))
= (Ax-4A + Bx+4B)/((x+4)(x-4))
= ((A+B)x + (-4A + 4B))/((x+4)(x-4))
Thus, equating it back to the original equation:
A+B = 0
A = -B
-4A + 4B = 32
-A + B = 8
2B = 8
B = 4 -> A = -4
Not too bad, actually. Now we have, overall:
= int_7^8 dx + (-int_7^8 4/(x+4)dx + int4/(x-4)dx)
= int_7^8 dx - int_7^8 4/(x+4)dx + int4/(x-4)dx
= [x]|_(7)^(8) - [4ln|x+4|]|_(7)^(8) + [4ln|x-4|]|_(7)^(8)
= (8-7) - (4ln12 - 4ln11) + (4ln4 - 4ln3)
~~ 1.8027 "u"
The exact answer is:
color(blue)(1 - 4(ln12 - ln11 - ln4 + ln3))
