It is readily seen, that since by this formula we may reduce the determination of fx" (a+b) to that of fx (a+b), we may also reduce this latter to that of fx”TMm x (a+bxTM), by writing n-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 f (a+ba), by means of fx-3 (a+b), and so on. In general, if r denote the number of reductions, we shall at last come to Jax (a+b), and the last formula will be n―rm+1 k+1 (a+bx) —a (n—rm+1) ƒxTM-TM ƒ (a+bxTM)* b { km +n+1—(r−1)m } From this formula it is evident that if n+1 be a multiple of m, then fr" (a+b) will be a finite algebraical quantity, for in that case the coefficient n-rm+10, and therefore the term containing fx-rm 'x (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" x (a+bxTM)"=ƒx” x (a+bxTM)^-1 (a+bx”)= n+m aƒx" x (a+bxTM)*-1 +bsx”+” i (a+bxm)*-', and that the formula (A), by changing n into n+m, and k into k—1, gives Substituting this value in the preceding equation, we obtain (B) n+1 fx" x (a+bx)= 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 f (a+b) to depend on that of farm x (a + bx, rm being the greatest multiple of m contained in n, and s the greatest whole num ber in k. The integral of fr1x (a+br3)3, for example, may be reduced by formula (A), successively to fix (a+br3), fri (a+br3); and by the formula (B) szx (a+bx3) is reduced first to sxx (a+bx3)3, and that again to sxx (a+bx3)3. 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 x-m+1 fxn- 'x (a+bxTM)k— a (n−m+1) Substitute n+m, in place of n, and it becomes (C) 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 Then writing k+1 in the place of k, we find (D) #+1 Sx” x (a+bxTM)*= k+1 (a+bx) +(n+km+m+1) sx2 x (a+bxTM) k+1 (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 Binomial Fluxions. (1—xx)1. (1). Reduce the integral of 10 (1-xx) to that of x (1-xx). 3 (2). Reduce the integral of xx (1—xx)3 to that of r (1—xxi. (3). Reduce the integration of S (1+xx) 3' to that of f 63. Let Pr be a rational fraction (P and Q being functions of x); and let the greatest exponent of x in P be less, at least by unity than in Q. If it be not so, divide the numerator by the denominator, till this latter condition takes place. For example, if the fraction 64. This done, find the factors of Q, as if we had to resolve the equation Qo; and if they are all of the first degree, real, and unequal, the proposed fraction will then be of the form To find the fluent in this case, we must decompose the proposed Ax Br Cx fraction into these others. +. + +&c.; of which the inx-fx- -g x-h tegral (art. 56) is Al (x-f) + Bl (x-g)+&c. (where I stands for the hyperbolic logarithm of the quantity within the parenthesis.) To this value we must add a constant C, and determine the coefficients A, B, C, &c. by first reducing them to the same denominator, and then transposing, and successively equating to zero the coefficient of each power of x, which process will give as many equations as there are unknown quantities. I decompose this fraction into the following ones: A+ Br ·+· x ax Cx a + x and reducing to the same denominator, transposing, and ordering the terms, I find Aa2+Bax+Bxx -1+Ca-A aa 2aa 1 + 2aa le Where re presents the correction C, the form of which is optional. αα 65. This method will always succeed when the factors of the denominator of the proposed fraction are all real and unequal; but it several among them were equal; if, for example, (r-a)" represented any number m of these factors, we must then decompose the fraction into the following ones. -g (x-a)TM and, after having determined the coefficients as before, we may pro ceed to integrate A'x- 1 (-a), by making x-a-z. Example. Let it be required to find the integral of (x3+x2+2) x ≈ (x−−1)2 (x+1)3 I suppose. + (x3+x2+2) x Ax (Bx+C) * (Dx+E) ż 7 whence A=2, B——3, C=1, D=−0, E=—1; ; therefore (5x+7) Now to inte(x+1)2 2x +1 (7—3x) x of which the fluent is ——-34z=31(x-1); and treating in the same manner the other fraction, we obtain for the complete integral or fluent Q there should be imaginary factors; on representing one of them by x+a+b√-1, there will be another of the form x + a−b √ —i, and their product x2 + 2ax + a + b will be a real factor of Q. We must therefore find the coefficients 2a, a, b, of the factor of the second degree, which is always possible; and the real factor of the second degree a2+2ax + a2 + b2, or more concisely 2+mx+n, will be determined. Then we must suppose that (Ar+B) +mx+n x is one of the partial factors of the proposed fraction, and determine A and B, as before. After which, making x+mz, the fraction will become A'z+B) ż 2 arc tang +C; and consequently we shall by this means arrive at = (1+≈) (1+≈≈) ̄1+z 1+22 B=-1, C==—, this changes the proposed fraction into these ; we shall find that A=, |