58. Hence we may conclude that any binomial fluxion represented by the formula 3" } (a + br")* is integrable algebraically. I. When n=m-1, whatever may be the values of m and k. II. Whenever k is a positive whole number for all values of m and n. We shall add two other cases in which an exact integration is possible. 1-1 (2-a) 59. Let a +6x" —; we shall have x +1 and taking n+1 b 1 m n+1 gives m 3 *« a 14 8 the fluxions on both sides ra e= xz (z–a) a fluxion n+1 mb admitting of integration whenever 1+1 is a positive whole number. For example, let the proposed fluxion be x3 is (a’ +r?)}, which –2. The transformed equation then becomes S1z3 z 3 1 sz:-)=3(2+2)/(a+2) = (a +2") (4x2—3a-). Consequently S7° : (a* +xm'= c + (a*+2014 (4x —Sa“). 60. r* 3 (a + bx")* = xmk +n i (6 + ar-my, as will appear by die viding the terms within the parenthesis by me, and multiplying the other factor by the same quantity ; but this fluxion is integrable from what we have just shewn, when mk +n +1 or-k~*+1) is a positive whole number. m m --2, and xmk to For example, let the flusion be. or as it may also be ** (a +2?) written, *8a+x2); we shall have mk+n+1 · mm (6+ax-me will be r *(1 + ax-5)-5. Now suppose I tax2, and we shall have for the transformed 1 *:(2-1); or 1 expression 1 2*, the 3 aa 2+ 3 aa 3 aa When a binomial fluxion has not the conditions which we have enumerated, we must endeavour to reduce it to some other known fluxion ; such, for example, as that of the quadrature of the circle, or of logarithms, &c. If this reduction is possible, it may be effected in the following manner. A method of reducing the integration of binomial fluxions to that of other known flurions. 61. This reduction will be effected upon the following principle. We have seen (art. 10) that flux uv = uv+vi, and conversely un= suv+fvi, or by transposition, juj=uv-svi. Therefore if we decompose the binomial 2" x (a+bx" jt into two factors, one of which is integrable, we shall cause the integration of the proposed binomial to depend upon that of the other factor, which in many cases is more simple than the given Auxion. This mode, which is at once extensive and curious, is called integration by parts. Among the different ways of resolving the proposed binomial into factors, we shall choose that which diminishes the index of 's without the parenthesis. This is effected by writing the proposed fluxion it = (a+bx")*. By this means the factor **-1 = (a + bxmJk is integrable, whatever be the value of k (art. 57). Now if we consider the factor omn+1 ==u, we shall have com1 = (a+bx*}'=ana (a+bum)*+1 . (k+1) mb therefore, since suv=uv-svi, we shall have fr" : (a+b")"= im+1 n-m+1 fitm-i(a+bx^j*+1 (k+1) mb (k+1) mb but fix* pa+bxmJk+1=Sx-oira+bxjk (a+bum) =a fixa (a+bum)* +bft" i (a +62m), therefore substituting this last value in the preceding equation, and collecting into one the terms involving the integral femmi (a+b.xm )*, we find nm+1 (1 + -).8x** (a + bum)" = (k+1) m k+1 mb (k+1) from which we get (A) Sæmi (a+b.x* )= k+1 br =v, and 3 ܀ It is readily seen, that since by this formula we may reduce the determination of fx" ; (a + bx")* to that of /ir«(a+bx"), we may also reduce this latter to that of fx-2: (a + bx)", by writing 1—m in the place of n, in the equation (A); then by changing n»m into n-2m, in this last equation, we shall be able to determine forms (a + bxrm), by means of fx*–5m (a + bx”), and so on. In general, if r denote the number of reductions, we shall at last come to J.xu-om : (a + bx), and the last formula will be Som17–1993 (a+bomye = k+1 6{km+n+1—(r–1)m} From this formula it is evident that if n +1 be a multiple of m, then fix** (a + bx*) * will be a finite algebraical quantity, for in that case the coefficient n –rm+1=0, and therefore the term containing frammi (a +b.xm)* will vanish. 62. There is another method of reduction by which the exponent of the quantity within the parenthesis may be diminished by unity; for this purpose it is sufficient to observe, that Sx" ; (a +bx*)*= fx" 3 (a +bxm)k-1 (a+bx^)= ntm afr" : (a + bx)k-1 + bfx (a+bum)k-1, and that the formula (Ă), by changing n into n+m, and k into k-1, gives mim (a+bx**-1 = b (km +n+1) Substituting this value in the preceding equation, we obtain (B) fix"(a + bux")* = n+1 (a + b.xml +kma frica +6m)-1 (km+n+1) By means of formula (B) we may take away successively from k, as many unities as it contains, and by the application of this formula, and also of formula (A) we may cause the integral fets (a+bum)k to depend on that of fænom s' (a + brought, rm being the greatest multiple of m contained in n, and s the greatest whole number in k. The integral of fx? ; (a+br}}}, for example, may be reduced by formula (A), successively to St* ; (a+bry, fië (a+bx+y!; and by T the formula (B) jzë (a+bx)is reduced first to sex (a+bx3 )}, and that again to Seč (a+bx+)? 63. It is evident that if n+1 and k were negative, the formula (A) and (B) would not answer the purpose for which they have been investigated. In that case they would increase the exponent of x without the parenthesis, as well as that of the parenthesis itself. If, however, we reverse them, we shall find that they then apply to the case under consideration. From formula (A) we deduce femmx (a+barm) *-m+1 (a+b=*)+2_6 (km+n+1) 3x6 (a+bx" } a (n=m+1) firi (a + 6.") ntm (a+bcm), a (n+1): a formula which diminishes the exponent of r without the parenthesis, since n+m becomes n+m, when we put -n in place of n. To reverse formula (B), we must take Sx" : (a + bx^)k-1= kma six" (a+b)' = k+1 k+1 (a +6x) +(n+km +m+1).S7" (a+bx^)" (k +1) ma This formula answers the end in view, since k +1 becomes -k+1, when k is negative. The formula (A), (B), (C), (D), cannot be applied when their denominators vanish.” This is the case with formula A, for example, when n+1=-km, but in all such cases the proposed function is integrable, either algebraically or by logarithms. Examples on the Reduction of Binonial Fluxions. (1). Reduce the integral of .x10 + (1-xx)* to that of 1 (1—31) (2). Reduce the integral of 3* + (1xx)to that of < (1-«o». (3). Reduce the integration of to that of s (1 + x)? 1 +13 nti x+ c |