Now, a+x+1 = x2+x+4+÷= (x+1)2+÷,

and, if x++=y, or x = y; then we have xdx = ydy➡dy.

= 4 { log (x−1)— log √aa+x+1+√⁄3 tan −1

IV. Let it be required to integrate du =

Ax+B (x2+a2)0

dx.

2x+1

√3

A

and u

dx

́ 2 (n−1) (22+a2)=1 + BS (x2+*+*a*«*

To obtain the integral of the differential

dx
(x2+a2)¤1

Kdr

we may assume

(1)

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

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

+

dx

(A)

.. x1+2x3+3x2+3 = Ax+B+ (Cx+D) (x2+1)+H (x2 + 1)2..... (1) Let x2+1=0, or x2 = hence eq. (1) becomes by substituting for Ax+B its value — 2x+1 x2+2x2+3x2+2x+2 = (Cx+D) (x2 + 1) + H (x2 + 1)2

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

By formula (A) we have, making n = 3, and n = 2

1

tan

+C.

(3.) — log(x-1)+9 log (x-2)-log (x-3)+C

(4.) ¦ log (x+1)— ¦
¦ log (x2+1) + 1/

(5.) – 1⁄2 log (x−1)— — log (x+2)+

(12.) —- log (x−1)—— log (z+1)+ log (x2+1)+C= log c√a^—1.

4

æ+1

tan

CHAPTER III.

ON THE INTEGRATION OF IRRATIONAL FUNCTIONS*

The first class of integrals which we shall consider are comprehended under the general form

x±” (a3 + x2)± dx

The integrals belonging to this class are, for the most part, obtained by the method of parts. Let us first take those of the form

The limits and nature of the present work will not permit us to enter at any length upon a sub. ject so extensive and intricate as the Integral Calculus. We shall therefore merely indicate the process by which some of the most useful integrals may be found, and refer the student who wishes to prosecute this subject to the masterly work of Lacroix.

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