477 √3 arc tan 22 √3 Consequently by substituting the value of z, we have for the complete integral l—2l (1+x)+§ 7 (1+x+xx)+; 1 1 arc tan 1+x √3 (2x+1) √3 +C. 67..The last case which remains to be considered, is that in which the denominator Q has one or more factors of this form (xx+ax+b)”. Here we must suppose that the partial fraction proceeding from this (Ax2m¬1 +Bx2n¬2+&c....+R), and then determine the factor is (xx+ax+b)" coefficients as before. After which, making x-z-a, and substituting, the fraction will become of the form A'zam+Bz +&c. +R' (zz+b'b')TM A'zam-1 z, which may be thus decomposed (zz+b'b') + B'zam-2 2+&c. any odd power of Now those terms in which the numerator contains z, are integrable, in part algebraically, and in part by logarithms (57); and those in the numerator of which z is raised to any even power, art (63); that is, we are able to integrate them, partly algebraically, and partly by circular arcs; consequently we shall by combining these means obtain the integral of the proposed fraction. 68. The following example will serve to elucidate these different methods, as it contains them all. Ar 1+x (Bx+C) x _ (Dx +E) ¿ _ (Fx3 +Gx2+Hx + I) ♬ XX + xx+2 + (xx+1)2 We shall find by reducing these fractions to the same denominator, that A=11, B=-—-—-, C=1, D=1, E=-1, F=1,G=-1 12 H=23,1=-2 and that the proposed fraction= (x+1)* and- · (xx+1)2 we must have recourse to the method of reduction before explained. Thus by formula (A), article 61, we find f 1+xx and again by formula (D) article 63, it appears (xx+1)22 8 1+xx 8 arc tan z; and arc tan r. Consequently by adding all these fluents together, we obtain x I 2x arc tan +C, the fluent of the fraction proposed. √2 69. From what precedes, it appears that every fluxion under the form of a rational fraction is integrable, either algebraically, or by logarithms or circular arcs. The only difficulty consists in finding the factors of the denominator Q. But this arises rather from the insufficiency of algebra, than of the methods of integration which we have given. Consequently when we can render any fluxional fraction rational, we shall be sure to obtain its fluent. We will now proceed to exhibit a few cases in which this is possible. 70. And first let us suppose that in the quantity proposed there are no other radicals than such as consist of a single term; (2√x+x√/x+xx) ample such as ~~√*+*+*x); I write it thus for ex x=212, x = 12 z11%, the fluxion will become rational, and consequently integrable. Again let X be a rational function of x; in order to find the integral of y Xx√ (a + bx + cxx), I find the two factors of a+bx+cxx; if they are real, I have √(a+bx+cxx)=√(m+nx) (p+qx). Supposing this quantity=(m+nx) z, and squaring it, we shall have p+qr=(m+nx) zz, whence -P-m 22, and n zz-q (pn-mq) z ̧ nzz-q ,י 2zk (mq-pn). Consequently (m+nx) z= (nza-q)2 √(a+bx+cx3). These values being substituted in the formula Xx √(a+bx+cr2) will render it rational, and consequently integrable. The same thing would evidently take place, if we were required to integrate the formula y = Xi EXAMPLE 1. Let y=(aa-x), make √(au-xx=(a−x) x; then a rational quantity of which the fluent is Ex. 2. Let y = √(xx—aa) ; making √(xx—aa)—(x—a) z, we y==24, and y=logo (2+1)—log = { *+√(xx—aa) shall have y= = a »} 71. When the factors of a+bx+cx' are imaginary, we must exterminate the second term of this quantity by supposing ≈+ b 2c and then Xx/(a+bx+cxx) becomes of the form Zz √(xx+b'b'). Let then (z+b'b')=z+u, and we shall have b'b'-uu 2u , and ➡(zz+b'b')=x+u= bb'+uu and z= u (b'b'+uu); these va 2uu lues substituted in the formula Zz (zz+b'b'), or 2u For example, let y ≈≈√(xx+da); making √(xx+aa)=x+2, we shall have y=xx+zi ; but i— 22 2 XC aa aal (x+√xx+aa) aa la. Now let C-aa-aa la -- C', and we shall have y = C'+ √(xx+aa)+} aa l (x + √xx+aa) 72. This same method might have been applied to the first case in which a+bx+ca2 has two real factors; for, on exterminating the second term, we shall have to integrate z√(zz-bb), or (bb-zz). And if we suppose √(zz-bb) = z—u, or √(bb—zz)—b—uz, we shall make both fluxions rational. 73. When a fluxion is not susceptible of an exact integration, we must have recourse to approximations, and series are then one of the last resources. In effect, it is easy to perceive that by reducing to a series any function X of the variable, we shall have a succession of simple terms, the aggregated fluents of which will give a near value of Xi. For example, we already know that the fluent of. a + x is the log |