% arc tan. 1 m1 i_ (2x+3) < + To integrate this last quantity, make (1+x)? '1 +x+xx *=2—1, and it becomes (3–4)ż_zz x of which the 2z+ zz+ 22+ 1 2z integral is } { (zz+1) 1 )- V3 1 plete integral l«—A (1+x)+31(1+x+xx)+ arc tan 1+* V3 (2x + 1) +C. V3 67. . The last case which remains to be considered, is that in which the denominator Q has one or more factors of this form (2x +ar+6)". Here we must suppose that the partial fraction proceeding from this factor is (A.x?m-+ Bx21? +&c.... +R) .x, and then determine the (xx+ax+b)" coefficients as before. After which, making x=-- a, and substituting, the fraction will become of the form A'zam-' + Bz + &c. +R 2, which may be thus decomposed A 227 Bizam- % + (zz+b6') (22+66) Now those terms in which the numerator contains any odd power of , 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, ; may be reduced to the form (zz +6%)m? 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. being of the form. Mz2 zz +67-by ; Let the fraction (1+x) xx (xx+2) (xx+1)2 • Because 1 2 z2+1 2 4z2+3 ✓3 4z2+1 V3 Azz 3 3 +1 1 that A , B=-1,C,D,E,F, 12 3 Aj (Bx+C) = (Dx+E) 3 (Fr+Gr’ +Hr+1): + 1+2 xx +2 (xx+1) We shall find by reducing these fractions to the same denominator, A CD н H=1 I= - and that the proposed fraction 4 (x-1)8_(4x4f**+¿x-1): + + 12 1+ 2 6 XX + 2 (**+1)' 1 1 1x 12 1 + x 12 1 I 1 2x 1 + XX Thens = ((1+4) = (). در tir xi l'ir+2) 12 1 1 12 12° xx+2 62xx+ 1 arc tan 672 72 23 2 1 4 (rx+1)* 8 xx+) 3 Now to integrate the partial fractions -5. and (*x+1) we must have recourse to the method of reduction before (2x+1) explained. Thus by formula (A), article 61, we find / (1+r)' ts and again by formula (D) article 63, it appears 1+xz (1 + r).' i thats =} +S: } (1 + r) 1+rr arc tan x. 8 1+xx 8 (x1+1) Consequently by adding all these fluents together, we obtain 1 1 1(1+x) + ST1+x2) x2 (+2) (x2 +1)=12 1(xx+2) + 12 1 (x+1) 1 1/3 $ 1(25+1) + 능 1+xx 62 arc tan 42+C, the fluent of the fraction proposed. 1 arc tan t 2 2x 2 (x-*+1): x+ 5. ? (x+1) (x+1) (+*+2x+3x2+3)=; 6. (x2+1)3 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 bave given Consequently when we can render any fluxional fraction rational, we shall be sure to obtain its fluent. We will now proceed to exhia bit 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; for ex(t+5Vx+13) & ; I write it thus ample such as st.IN ; I x=2", r = 12" 2, the Auxion will become rational, and consequently integrable. Again let X be a rational function of x; in order to find the integral of j=Xin (a + bx + cxx), I find the two factors of a+bx+cxr; if they are real, I have v(a+bx+crr)=v(m + r) (P+qx). Supposing this quantity=(m+nz) 2, and squaring it, we shall havep+qr=m+11x) zz, whence ==P-M 22, and ; ( = n zz-9 nzz-9 aze (mq—pn). Consequently (m+na) z=(on-mg) == . x (n23-9) (a+bx+cx*). These values being substituted in the formula Xiv(a+bx+cra) will render it rational, and consequently integrable. The same thing would evidently take place, if we were Xi required to integrate the formula y v(a+bx+cr?) EXAMPLE 1. Let y=iv(aa—31), make v (aa-rx=(a-r) z; then == =v(aa—); 1+zz (1+zz)?’ 8a? gai consequently y= a rational quantity of which the fluent is (1+z2)3' -a +az 2az 1 + zz easily found. shall have y=and y=log. (**)=log:{*+V(+3–a)} ) Ex. 2. Let y= ; making ✓(tr-aa)=(5-a) 2, we (xx-aa) = c+1 2_1' 71. When the factors of a +bx+cre are imaginary, we must ex 6 terminate the second term of this quantity by supposing :+ and then Xin(a+bx+cxx) becomes of the form Z(x2+663. Let then w(x+6)=2+4, and we shall have bb—uu -2, and 2u V(x+6)=x+u= u 2uu lues substituted in the formula Zi (x+66'), or 'zi will V(x+b%) render it rational. z aa 2 aa For example, let y = x1(**+da); making v(re+aa)=x+%, we shall have y=xi+zë ; but i=-(aa+zz) therefore y=xan := xiand y=C+11x42 2-z=c4aa+2V (2x+aa) + C2 faal (x+ 1 xx +aa) - da la. Now let C-4 aa-aa la -- C', and we shall have y = C++(xx+aa) + } aal(x+ Vxx+aa) 'V ' 72. This same method might have been applied to the first case in which a +bx+cx? has two real factors; for, on exterminating the second term, we shall have to integrate z (zz—66), or zv(66—22). And if we suppose (zz-bb) = 2-U, or (66—zz)=b-uz, we shall make both fluxions rational. 73. When a fuxion 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 $X.r. For example, we already know that the Auent of is the log atx |