210 
MR. TALBOT’S RESEARCHES IN THE INTEGRAL CALCULUS. 
And by our method, 
/* x Q dx 2 4 
S / , == = — V + 
J V 1 — X s 
const. 
it remains to verify that 
x/l — x 3 + n/ 1 ~ 3/ 3 + \/ 1 — z 3 = — v* • 
In order to demonstrate this, let us resume the original equation x 3 -\-vx 2 — 1 =0, 
which gives rx 2 = 1 — x 3 , 
\A> . x — s / 1 — x 3 , , 
and similarly, 
\fv • y = \/T - j/ 3 
v . z = *y i — s 3 
.‘. \/l — \/! — 3/ 3 + — * 3 = \A> (a? + j/ + a) 
= \/ v ( — v) — — V s . Q.E.D. 
. /» .i’ 
•J V {X s + A’ 2 ) + y'.r 3 + Jt‘ 2 * 
This is a function of the binomial x 3 + x 2 , which being put = i;, vve have 
x 3 + x 2 — v = 0. 
The three roots of this equation are the variables that answer the problem. 
Putting \/v + v = <p v, the sum of the three integrals becomes 
/£+/£+/£=/£ Srf,, 
but S dx = 0, because x + y + * = — 1 3 
.*. sum of integrals —J 0 = const. 
Ex. (i. J dx s/\ + x \ 
Here we cannot suppose \/l x n = v a symmetrical quantity, because that would 
amount to the supposition v 1 + x n = \/ 1 + y n — \/ 1 + z n = &c., which implies 
that x = y — z = &c., whereas we suppose the roots to be in general all different from 
one another. This is the reason why it was remarked before, that it was requisite the 
polynomial should contain at least two distinct powers of x. When that is not the 
case, a second power of x must be introduced. There are several ways of doing this ; 
the simplest is the following : 
Put f (lx j _|_ x n under the form f* x dx y/ 1 -. Assume 1 = v, 
.’. x n — v x 2 + 1 = 0 ; 
an equation of/? dimensions, of which v is the only variable coefficient. The n roots 
of t his equation answer the problem. 
