MR. TALBOT’S RESEARCHES IN THE INTEGRAL CALCULUS. 
17 
/. x 3 x — r = 0, 
and 
Therefore 
blit 
V 1 + x 4 2 x l 
_ S i/l | ■ y4 . dx — - — + S^f, 
~ d x dr 
as in the last example ; and 
S x 2 
— — S x dx — — dr 
r 
- S 
1 + 
X 3 
d r 
^.dx = ^-dr = -~dr 
4 4 
•• s / 
^ 1+x \dx = ^-r + C. 
x * 4 
This result therefore agrees with the last example, and gives the same theorem, but 
it supplies a different demonstration of it. 
We will now suppose 
V'l + x 4 , , a , v 
o * A I "l 
X X x z 
a being a constant. This gives 
cd + 2 v 
X 3 + 
2 a 
x 2 + v x + 
u 2 — l 
— 0, 
and I find this result, that if three abscissae are roots of this equation, which may be 
written 
x 3 — p x 2 -f- q x — r — 0, 
then the sum of the arcs 
*0^ 
= p — — + const. = <p v + C. 
This sum is therefore constant if <p v is so. 
Let v=k,v — Jc', be two values of v, which give the same value to <p v, or p — — . 
Let the three hyperbolic arcs in the first case be a a' a", and in the second case 
|3 /3' /3", then 
a + a' + a" = (3 + (3' + j3". 
All the abscissae have the same origin at the centre of the curve, therefore the arcs 
have the same origin, and therefore can be subtracted from one another. Therefore 
putting cc — (3 = y, a! — /3' = y', a" — |8" = y", we have 
y ~|~ y f y" = 0. 
This appears to me to show the possibility of finding three arcs such that (neglecting 
MDCCCXXXVII. D 
