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(28+1)3=+=

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3

3 s

arc tan x. (*x+1) 8 1+xx 8 Consequently by adding all these fluents together, we obtain

1

1

1 (xx+2)+

1

(1+x1) ** (**+2) (*2 +133= 12 ?(1+2) + ?(3+2)+ 1/ 12(**+1) + arc tan 52+C, the fluent of the fraction proposed.

1

1 2

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1 23

} (x+1)

1+xx

arc tan I

1 672

2

Examples for Practice.

1. Required the fluent of

8 + x?—x4 (2—4x) <

( 2.

XP-X-2
(x + 2+2)

+, 3.

42. + xx,

(x-x+1) 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 example such as ()*+:$Vx+13) I write it thus

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nzz-9

-a +az?

x=2!?, : = 12 zo2, the fluxion will become rational, and consequently integrable.

Again let X be a rational function of x; in order to find the integral of j=X&v (a + bx + crx), I find the two factors of a+bx+cxx; if they are real, I have v(a+bx+crr)=v(m+n) (p+qx). Supposing this quantity=(m+nx) z, and squaring it, we shall have p+qr=(m+nx) zz, whence :=PM 22, and ;=

-9 2zx (mq-pn)

Consequently (m+na) 2–(pnmq) z (nz3-9) w(a+bx+c8*). These values being substituted in the formula Xi (a+bx+cra) will render it rational, and consequently integrable. The same thing would evidently take place, if we were required to integrate the formula y

v(a+bx+cr?) EXAMPLE 1. Let y=iv (aa—-1), make (aa-rr=(a-) z; then

i=
Hazż
and (amr)2=

2az

=v(ad); 1+z2 (1 + zz)??

1 + zz 8a za i consequently y=

a rational quantity of which the fluent is

(1+zx)* easily found. Ex. 2. Let j =

; making (tt-aa)=(:-a) 2, we V(xx-aa)

-22
y
an

c(2+1) 2_1' 71. When the factors of a+bx+cxare imaginary, we must exterminate the second term of this quantity by supposing ++

b and then Xin(a+bx+csx) becomes of the form Zz v (2x+bb). Let then v(x+66)=x+u, and we shall have..

2u bbt uu

ü (x+669=3+4= and 2= (66' + uu); these va

2u

2uu lues substituted in the formula Zi v(x+66'), or

zi

will

(x+6*6') render it rational.

+a)}

6'6'-uu

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For example, let y = x1(**+da); making (r* +aa)=x+2, we shall have y=x2+zi; but i-- (aa+zz) therefore y=ximaa.

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2

2

and y=C+£12–21-1*=C-aa+ V (xx+aa) + aal (x + 7 xx +aa) aa la. Now let C-aa-aa la -- C, and we shall have y =C+V(xx+aa) + aal(r+ Vär+aa)

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 v(zz-—66), or 27 (66—22). And if we suppose (z2–66) =%, or (66—22)=b-uz, we shall make both fuxions rational,

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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 SX1.

is the log For example, we already know that the fluent of

at*

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a

+

a

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-&c. ;

(a +x). Now if we reduce into a series, either oy actual di.

ati vision or by the method of indeterminate coefficients, art. 41, page 95, we shall find

+ -&c. Therefore or log Q+x

at de 203 (a+x)=

&c. +C. If we make x=0, the constant

2u? 3x3 C=log a ; consequently we shall have l (a+x)= la+* 23 and writing

- for x, we obtain l (ams) = la
3a'
?

az!
+
+&c.). Suppose now that it

a+2 substitution we shall find I (am) = 2la-1 (a+2) =la

atz -&c.; and therefore I (a+z) =la+aiz+zca+z)

+ 2 (a+z)? &c., a series converging with a degree of rapidity proportionate to the excess of a above 2.

For example l 100=l (99+1)=199+ +&

:+1 1

+&c. -2.397, &c.* 2.112

(1+

or x =

2a

3a2

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+

Z

1 100

or arc tan irt

If we had y = then

y=arc tan x; but. being re1+XX

1+IX duced into a series, gives y=:~ri+z*=~~°*+&c. · therefore y

1

ms x? + &c.

5 7 Let now

y
be

any arc, x its sine, or y=arc sin t, we shall have j=

1-xx); this expression expanded into a se(1-4X)

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23 +

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The logarithms found above are the hyperbolicor Naperean ; these are directly converted into the common logarithms, by multiylying them by the constant factor 0.434294482.

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1

+

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1.3.5 ries by the binomial theorem gives ģ=i (1+1+*+1-3

2.4

2.4.6

1 23.1.3 25 1.3.5 +&c.). Therefore y or arc sin x=x+ +

+ 2 3

2.4

5 2.4.6 7 +&c., a fluent to which it is not necessary to add a correction. Letx=1, and call the semicircumference , and we shall obtain =1

2 1 1.3 1 1.3.5 1 + +

+ &c. If x= 3 2.4

5 2.4.6 7

1 1.3 1 1.3.5 1 1+

+&c. 3.23 2.4.5.25 2.4.6' 7.27 74. These examples suffice to explain the preceding method. That which follows will merit attention.

The formula flux (xy)=ay+yë, gives xy=fxy +Syř ; therefore, in general sxy=xy--Syx, and if we denote by X any function of t, we shall in like manner have /Xi=Xx-fxX. Call Å=X';; then by the same principle Sxx, or SX'x=X' **

2

X' x3
X = X", and we shall have X"-9***, &c.

2 2.3

2.3 Substituting these different values in the first expression, we shall obtain

as X' X"

-X"-&c. 2

2.3 2.3.4 2.3.4.5 or, supposing é constant

** +

2.2 2.3.32 2.3.4.23 EXAMPLE. Let X=1 and we shall have

at

Sxx Š. Let again

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2

X" +

+&c.

-1

2

2.8

&c. (a+u ja tott

ra+t)s' (a+*)* Therefore s

+&C.... + ato at*(++7(+)* that is

+

&c. as we havo before found its

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