8 
MR. TALBOT’S RESEARCHES IN THE INTEGRAL CALCULUS. 
braic quantity, which is considered to be impossible. This is verified by calculation ; 
for s " 
2 arc x = x */ 1 + x 2 + log (x + J 1 + x 2 ). 
Calling log (x + J l + x 2 ) —f. x, we have 
fx = 0-881372 / .y = 2-143099 
fz = 1*261722 fx + fz = 2-143094 
2-143094 sum = 0 - 000005 
This sum approaches zero very nearly. The quantities fx,fz are subtractive, being 
parts of 2 arc x and 2 arc z, which have been already shown to be so. 
Demonstration of Theorem II. 
Let v \/ 1 + x 2 = n x 2 + x + v, where n is a constant, 
■ • ^ n ' 0 -r — u, 
and 
S^/l -{-x 2 .dx = nSx 2 dx-\-Sxd 
x. 
— 2 . 
the term rS dx being omitted ; because, since S x = — is constant, the factor 
S dx = 0. 
The formula S x? — p 3 — 3 p q 3 r gives 
S .r 3 p 3 
~T = T~ PI + r > 
/ . 2 \ 2 
.*. (observing that p is constant and = — —) S x 2 d x — — d q -f- dr. Therefor 
the first term, or 
n S x 2 d, x = 2 d q n d r. 
The formula 8 x 2 — p 2 — 2 q gives the second term, or 
S x d x — — d q 
.-. »S x 2 d x + S x d x = d q -f- n dr, 
v S l + x 2 . d x — dq + n d r. 
Now we have (omitting constants). 
-e 
or 
V S' 
n ir 
and 
, 2 7 2 v dv 
; . dq — — dv ~ — , 
i n nr 
n dr = dv 
n 
d q n d r ~ — 
2 v dv 
