25. Er. 31. In a similar manner may be found the dividing till the remainder be which can always be done, and the fluent of that remainder will be had by the 8th form in vol. 2. Thus, by dividing first by x" + a", the terms are, — 1 — 1 j -222-I a2 xc n − 2 n − 1 * + a2n xen. -3n-1% -+ &c till the last term be ad−1)" (cd)n-1, and the remainder d)n-1 x - 1 when disc-1, or 1 less than c, which is also the number of the terms in the quotient; and therefore the fluent is has all the same terms hyp. log. of "+a". In like manner, Ex. 32. as the former, but their signs all+ or positive, and the re 1 mainder = a(c-1)" × hyp. log. of x” — a”. Also in like n Ex. 33. The fluent of manner has all the very same terms, the same with the preceding, by substitut. tiplying the whole series by the fraction → is also 26. When the numerator is compound, as well as the denominator, the expression may, in a similar manner by division, be reduced to like terms admitting of finite fluents. Thus, for Ex 27. There are certain methods of finding fluents one from another, or of deducing the fluent of a proposed fluxion from another fluent previously known or found. There are hardly any general rules however that will suit all cases; but they mostly consist in assuming some quantity y in the form of a rectangle or product of two factors, which are such, that the one of them drawn into the fluxion of the other may be of the form of the proposed fluxion; then taking the fluxion of the assumed rectangle, there will thence be deduced a value of the proposed fluxion in terms that will often admit of finite fluents. The manner in such cases will better appear from the following examples. Ex. 1. To find the fluent of √(x2+ a2)° Here it is obvious that if y be assumed = x/(x2 + a2), then one part of the fluxion of this product, viz, x × flux. of √(x2 + a2), will be of the same form as the fluxion proposed. Putting theref. the assumed rectangle y=x√ (x2+a2) into fluxions, it is j(r+a2) + But as the √(x2 + a3)' former part, viz, √(x2 + a2), does not agree with any of our preceding forms, which have been integrated, multiply it by √(x2 + a2), and subscribe the same as a denominator to the product, by which that part becomes x = √(+); this united with the former part, √(x2+a2) + a'i √(x2+a2) √(x2+ a2) ; hence the given and its fluent is hyp. log. of x +√(x2 + a2), by the 12th form of fluents. found from that of) by the same 12th form, and is = {x√/ (x2 — a2)+ax hyp. log. x √(x2 — a2). Ex. 3. Also in a similar manner, by the 13th form, the fluent fluent of will be found from that of and √(a2-12) √(a2-12) comes out • 1x √(a2 x2)+ax cir. arc to radius a and sine x. Ex. 4. In like manner, the fluent of (+) will be a2) Here it is manifest that y found from that of √(x2 + a2)° must be assumed = x3√(x2 + a2), in order that one part of its fluxion, viz, x × flux. of √(x2 + a2) may agree with the proposed fluxion. Thus, by taking the fluxion, and reducing as before, the fluent of will be found = 1.x13 √(x2 + a2) — 3a2 × ƒ Ex. 5. Thus also the fluent of √(x2 + α2) √(x2 is x(x2-a2) √(a2 - x2) is — x3 √ (a2 In like manner the student may find the fluents of √(x2±a2)? √(x2 + a2)3 number, each from the fluent of that which immediately precedes it in the series, by substituting for y as before. 28. In like manner we may proceed for the series of similar expressions where the index of the power of x in the numerator is some odd number. Ex. 1. To find the fluent of y = x2√(x2 + a2), and taking the fluxion, one part of it will be similar to the fluxion proposed. Thus, y = 2xx x3x √(x2 + a2) + √(x2 + a2); hence at once the given fluxion x3% √(x2 + a2) =; 2xx(x2+a2); theref. the required fluent is yf. 2xx√(x2 √(x2 - ƒ . 2xx √(x2 + a2) = x2 √√/ (x2 + a2) — 3(x2 + a2)3 ̧ by the 2d form of fluents. Ex. 2. In like manner the fluent of ¿¿12 u̸/ (-212 ——— α2) — }(·x2 — a2)3. · √(x2 + a2)? Here it is manifest we must assume y = This in fluxions and reduced gives j = √(x2 + a2) √(x2 + a2). 5x5x + = }j 5 √(x2 + a3) is y — a2 × ƒ √(x2 + a3) = {x2 √/ (.x2 + a2) −‡a3×ƒthe fluent of the latter part being as in ex. 1, above. In like manner the student may find the fluents of where n is any odd number, viz, always by means of the Bluent of each preceding term in the series. 29. In a similar manner may the process bé for the fluents of the series of fluxions, √(ax) √(a + 1)' √(a + x)' √(att)" Fusing the fluent of each preceding term in the series, as a part of the next term, and knowing that the fluent of the is given, by the 2d form of fluents, = first term Ex. 1. To find the fluent of of √(x+4) xx √(x+a)' having given that = 2√√(x + a) = A suppose. Here it is evident we must assume y=x(x+a), for then its flux. j = +x√(x+a)=· + + √(x+a) + √(x+α) √(2+ a) √(x+a) =j-zaa; and the required fluent is 3y Za▲ = 3x √(x + a) - ja√(x+a)=(x − 2a) × 3√/ (x+a). In like manner the student will find the fluents of = B. Here y must be assumed = x2 √(x+α); of for then taking the flu. and reducing, there is found (x+a) 户 =zy - ав =2x2√(x + a) √(s+a) = 3 y - ‡AB = }x2√(x + a) – ‡a(x − 2a) × } √(x + a) = (9.212 -4ax +8a2) × 73 √(x + a). In the same manner the student will find the fluents of And in general, the fluent of √(x+a) 2 = being given = c, he will find the fluent of √(x+a) 2n+1 2n 2n+1 -ac. 30. In a similar way we might proceed to find the fluents of other classes of fluxions by means of other fluents in the table of forms in volume 2; as, for instance, such as xx(dx-x2), x2*✅ (dx − x2), x3x√√(dx − x2), &c, depending on the fluent of x/(dx-x2), the fluent of which, by the 16th tabular form, is the circular semisegment to diameter d and versed sine x, or the half or trilineal segment contained by an arc with its right sine and versed sine, the diameter being d. Ex. 1. Putting then said semiseg. or fluent of (dx—x2) =A, to find the fluent of xx√(dx - x2). Here assuming y = (dx − x2), and taking the fluxions, they are, j = 3⁄4 (dx — 2xx)√√(dx-x); hence x*(dx = x2) = ‡ d÷√ (dx − x2) — } j = då – }; therefore the required fluent, fxx√ (dx − x2), is i̟da—‡y =÷da—+(dx-x2)=в suppose. Ex. 2. To find the fluent of x2x (dx − x2), having that of xx✓✅ (dx—x2) given B. Here assuming y=x(dx − x2), then taking the fluxions, and reducing, there results = ({dxx - 4x2x) √ (dx − x2); hence x2+(dx − x2) = ždxx √ (dx—x2)—‡j={dB-4, the flu. theref. of x2/(dx-x2) is {dB — ‡ y = {dB — ‡x(dx — x2)3. -I Ex. 3. In the same manner the series may be continued to any extent; so that in general, the flu. of -1√ (dx − x2) being given = c, then the next, or the flu. of x" will be - idc-x-(dx— x2)}. 31. To find the fluent of such expressions as a case not included in the table of forms in vol. 2. - (dx − x2) Put the proposed radical (± 2ax)=z, or x2+2ax z2; then, completing the square, x2± 2ax+a2 = x2+a2, and the root is x ± a = √(x2 + a2). The fluxion of this is |