le tle: es ae 
115 
peg eter OMe 
(10.) da =’ (A) da (a— 22) 
, Ian 2cA+b ch? + bA+ ca cA? + bA+ ca .. 
(11.) { F(a, y) dv=2f F( Oe, J. eth mace 
(12.) ue { F (x, y) dx= fr (2) da. 
In particular: 
a 
ae 
fs ‘dx dd. 
J V ax? + bx + ¢ 2) 
This method always furnishes the substitution by which the reduction is 
_ 2cA-+-b ; 
effected, viz.: 7 — 
A* —a 
Consider next the integral: 
fF, y) dx 
TT Vax? + be? + ex +d. 
(1.) y? = ax> + bx? + ex d. 
(2.) ==, Gi== 0. 
(3.) p+q=o. 
(4.) y => (2, 4) 
(5. ) 2 (a, A)—(ax3 + bx2+ ex +4) =0. 
__ o? (x, A)—(ax* + bx?+ ex 4+ d) 
ma iF =09, 
Ui" be 
(6.)  X(2, 4) 
Now the equation : 
X (m A)==o0. 
is of the second degree in x and hence can not in general be solved by rational 
processes. The necessary and sufficient condition for the reduction of the given 
integral to a rational integral is that the coordinates of at least one point of inter- 
section of L and the cubic: 
fs =y? — (ax® + br? + er +d) =0. 
be expressible rationally in terms of 2.. This will certainly be true if the cubic, 
f;=0, has a double point. For taking A at the double point two solutions of the 
equation : ; 
@? (xd) — (ax® + br? + ex +d) =0. 
are known, and, after dividing by (2 — a )*, the equation : 
xX (a A) = 0. 
is a linear equation, and can be solved by rational processes. 
