202
CALCULUS
can be referred to an integral already treated in § 2 by means of
the transformation
x — c = z 2 or = — z 2 , 0 ^ z,
according as the values of x between x 0 and x 4 make x — c positive
or negative.
Example. Consider the integral:
i
j V— (x
dx
(x — l)(x — 2)(x — 3)
Let x — 1 = — z 2 . The integral thus goes over into
o 1
' T —2 dz o r dz
J V(1 + z 2 )(2 + z 2 ) J V(1 + z 2 )(2 + z 2 )
The substitutions x — 2 = — z 2 and x — 3 = — z 2 would have led
to other forms equally tractable.
dx
4. The General Case,
j-
V Gi(x)
Let (r 4 (a,-) be a polynomial of the 4tli degree, whose roots or factors
are all distinct. If Cr 4 (x) has a real root, x = a, the transformation
(1)
y =
x = a + -,
y
will carry the integral into an integral of the form treated in § 3,
namely:
cly
/:
V Gs(y)
It remains, therefore, merely to discuss the case that
(2) G\ (x) = (x 2 + p t x + <p) (x 1 + p 2 x + q 2 ),
0 < 4 g 4 — pf, 0<4 q,-pl,
the second factor not being identical with the first. Let
(3) x ;= y + h.
Then x 2 +p l x + q l = y 2 +p[y + q[, x 2 +p 2 x + q 2 =y 2 +p! 2 y + q! 2 ,
where q[ = h 2 + pji + q u q'i = № + + Qi •
Let us seek to determine h so that q[ and q 2 will be equal: