first of these differentials; hence by (1) we have

dx x2+x

1

a

dy

dy

(ay2+by+c)*

(a y2 + by + c)1

·log {2 ay+b+2a* (ay2+by+c)*}by (6)

log

log

[ 2 a + b x + 2 a31 (a + bx + c x2)

х

2 a + b x + 2 a* (a + bx + c x2) +

EXAMPIES.

log

2x+1 - 5

92. The formula to be employed in integrating by parts is, Art. 86 (24) fudv = uv - Sv du,

where it is evident that fu dv will be known, provided fv du can be found. As this method of integration is extensively used in the Integral Calculus, the student should make himself familiar with its application, and endeavour to acquire the power of writing down results without repeating the formula.

. ' . ƒ x log x d x = ƒ u dv = uv – ƒ v du = 2

lcg x —

2. Let it be required to integrate (1 + x2)3 x3 dx*.

Let x = u, and (1 + x2)3 x dx

v = f (1 + x2)*;

= dv;

then du = 4 x d x and

•• . ƒ (1 + x2)3 x3 d x = fu dv=uv -
- fvdu

= ↓ x* (1 + x2)* − ƒ ƒ (1 + x2)* x3 d x .... (1). The integral in the last term of (1) which we have still to obtain,

This example has been chosen for illustrating the principle of integration by parts, but its integral may be more readily obtained by expanding (1 + x2)3. Thus S(1+x2)3 x3 d x = Sx3 d x + 3 S x2 d x + 3 § xo d x + S x11 d x 32-8 3210 x-12 2.6

+ =
10 12.
120

(10x636x + 45 x2 + 20),

1

differing only from the former result by the constant

120°

x2

must be submitted to the same process of integration. Thus let
x2=u, and (1 + x2)*x dx = dv;
then f(1 + x2)*x3 d x = ƒ x2. (1 + x2)* x d x

4

= хе

(1 + x2) 5

‡ƒ (1 + x2)3 x d x

x2 (1 + x®)° _ (1 + x2)° ̧

10

3. Let it be required to integrate ƒ (x2 + a2)* dx.

60

(x2 + a2) ‡ '

・(x2 + a2) +

dx

x2 d

· · · f(x2 + ao)* dx = a*

=

a2 log { x + (x2 + a2)2 } + S x

= a2 log { x + (x2 + ao) * } + x (x2 + ao)* − ƒ (x2 + a2)* dx.

Transposing the last term, and dividing by 2, gives

ƒ (x2 + ao)* dx = 2 log { x + (xa + a®)* } +

93. A rational fraction is a fraction of the form

(xTM + a xTM-1 + b xTM - 2 + . . .) d x
x” + a' x”-1 + b' x2-2 + ·

where the indices of x are all positive integers.

If the numerator contains powers of x as high or higher than the highest in the denominator, the fraction can always be reduced by common algebraic division to a mixed quantity, composed of a rational integral function, and a rational fraction in which the index of the highest power of x in the numerator of the differential is less by one or more units than that of the highest power in its denominator. We have only then to consider rational fractions in this case, and as they are not often met with in practice containing high powers of x, we shall confine our investigations to such only as are likely to occur in the most useful inquiries.

Rational fractions are integrated by resolving them into the sum of a number of fractions with simpler denominators, called partial fractions, as in the following cases:

I. When the simple factors of the denominator are real and unequal.

In this case, either of the following methods may be employed:
First method.--Let the proposed fraction be

A(x-b) (x-c) + B (x−a) (x−c) + C(x− a) (x−b).

Now this equation must hold for all values of x; hence if

Consequently A, B, C, are all known, and thence

(b − a) (bc)'

(c

Bdx

с

b)

('C dx

x = b + ]

x-c

(x − b) + C log (x − c)
(x − b)3 (x − c)° } .

a)

U fx A

=

=

Fx x-α

V

fx = A.

Fx x-a

+ B

+ C

Now since Fx = · (x − a) (x —b) (x−c), it is evident that when x =

to determine A we have only to divide fx by

in the result. Similarly, it is shown that B and C are determined by the same formula, writing in it b and c for x respectively.

u = — log (x-1)+log (x-2)++ log (x-3)= log C

(x − 2) (x−3)*.

By the first method we get from the assumed equation 2x5A (x − 2) (x − 3) + B (x − − If x 1, then 3 = A (1 − 2) (1

- 1 B (21) (2 3) = − B ... B = 1,
1 = C(3-1) (32) = 2C...C=;

==

the same values of A, B, C, as were obtained by the former method.