m

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

n

√a2x2+C.

* In this manner a certain 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

Р

Q

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

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

=22+a2, if

p2

=2+9=12 = 2+22, il a = √√√9-22;

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

2 {log. (a+x) — log. (a—x)+log. C} = L. log. Ca+x

=

2a

2a

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

a-x

Here -2x3+ x = x (x2—-2x2 + 1) = x (x2—1)2 = x (x+1)2 (x−1);

which, reduced to a common denominator, gives the equation

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

Substitute these values of A, B, D, in the above equation, and we have x3+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, · 2x2 (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)(x+x+1) x-1 x2+x+1

.. x = A (x2+x+1)+Bx (x−1)+C (x−1)
= (A+B) x2+(A−B+C) x+A-C;

hence, by the method of undetermined coefficients, we have

Now, x2+x+1 = x2 + x + 4 + z = (x+1)2 + 4,

and, if x+y, or x = y· ; then we have xdx = ydy -- 1⁄2 dy.

Differentiate this equation, and reduce to a common denominator; then,

1 = {(H+K) — 2 (n−1) H} x2+(H+K) a2

.•. (H+K) a2 = 1, and 2 (n−1) H = H+K;

=

2a2 (n−1) (x2 +a2)"~1 + 2a2 (n—)

a formula which, by successive operations, diminishes the exponent n, and

x+2x+3x2+3 Ax+B Cx+D

(+1) (ux2+1)= x2+1

.. x2+2x3+3x2+3 = Ax+B+ (Cx+D) (x2+1)+H (x2 + 1)2. . . . . (1) Let x2+1= 0, or x2 = - 1; then we have 1 − 2x = Ax+B; hence eq. (1) becomes by substituting for Ax+B its value −2x+1 x+2x+3x2+2x+2 = (Cx+D) (x2+1)+H (x2+1)2

or (x2+1)(x2+2x+2) = (Cx+D) (x2+1)+H (x2+1)2
.. x2+2x+2

=Cx+D+H (x2+1). . .

Let x2+1 = 0; then, as before, we have 2x+1 = Cx+D;

and this, substituted in (2), gives x2+1 = H (x2+1), or H = 1.

(2)

dx

+

+

(x2+1)3

2

(x2+1) 2