And in a similar manner, from the chapter on Inverse Functions, we have

a2-b2x2
dx
a2+b2x2

b

a

In all these integrals the radius of the arcs is unity, and the arbitrary con stant is not annexed to the integrals in the right hand column, for want of breadth of page. As it is frequently desirable to integrate differentials in which the radius is a instead of unity, we shall exhibit a few of those which most frequently occur to that radius. In the left hand column of the above differentials, write for 2, and we have to radius a.

a

These are the elementary forms to which every differential whose integral is required must be decomposed; and the reduction of expressions to one or more of these fundamental formulæ is the object of almost every process in the Integral Calculus. The following are a few examples.

m

= (v3 — 2a2 v + a*v ̄ ̄) dv

a=fde-20° fv dv+af

= · a2 v2 + a1log. v

=

=

v2 (v2—4a2) + aa log. v

4

(a2+x) (x2-3a)

4

to

= x2 (a2—x2)—+dx

Here =2, and the differential can easily be integrated.

n

In this manner a certair class of integrals may be very elegantly obtained.

CHAPTER II.

RATIONAL FRACTIONS.

It is readily proved, by the theory of equations, that every rational fraction

P

Q

of the form may be decomposed into others, which must have one of the following forms:

x-a' (x-a)"' x2+px+q' (x2+px+q)"

()

where the quantities A, B, p, q, n, .... are constants, and the factors of the expression x2+pa+q are imaginary.

Hence the last two of the forms in (1) are reduced to

Here du A (x-a)—"dx; therefore, by integration, we have

Here -2x+= ∞ (x1—2x2+1) = x (x2—1)2 = x (x+1)2 (x−1)3,

and, therefore, we assume

=

+

x=2x+x x

B
(x+1)2

C

D

E

+ (x+1) + (x−1)2 + (x=1);

which, reduced to a common denominator, gives the equation

x2+x2+2=A (x2-1)2 + Bx (x-1)+Cx (x+1)(x-1)2+Dx (x+1)2+ Ex (x+1)2 (x−1). A

Let x =

0; then 2 =

Substitute these values of A, B, D, in the above equation, and we have x2+x2+2=2(x2-1)2 - 4 x (x-1)2 +x (x+1)2 + Cx (x+1)(x−1)2+ Ex (x+1)(x-1),

or, 2x (x2-1)+ x (x2-1) = Cx (x+1)(x−1)2+Ex (x+1)2 (x−1) and, dividing both sides by x (x+1) (x−1), we have

=

+

(x-1) (x2+x+1) x-1 22+x+1

.. x = A (x2+x+1)+ Bx (x−1)+C (x-1)

=(A+B) + (A-B+C) x+A-C;

bence, by the method of undetermined coefficients, we have