By a process analogous to the above we can find the integrals of

When m is negative, the integral may always be found by assuming y

= √c (log. (2cx+b+2 √c√a + bx + cx2) — log. 2c}

ON EXPONENTIAL FUNCTIONS.

1o. If X = ƒ (a), then the function Xdx, if we make a2 = u wil become f(u) du

dX

2o. Differentiating Xe, we have e1 dx (X+dx) so that every exponential function in which the factor of e* dx is composed of two parts, one of which is the first differential co-efficient of the other, will be easily integrated. For example

In like manner, if we make 1 + xz, we shall find

In every other case, however, we must have recourse to the method of inte gration by parts.

Treating a 2-1 dx, &c. in the same manner, we shall finally have

It is manifest that the same method is applicable to Xa dr, where X is any

entire algebraical function of x.

But if the exponent n be negative, it is manifest that the exponent of æ must go on increasing; and therefore, in the integration by parts we must consider a as constant in the first instance, in this manner, if

a dx
24-1

Integrating in the same manner, we shall finally have

We cannot, however, proceed with our calculation beyond this point, because we should obtain a result = α

We can, however, approximate to it in the following manner

If n is fractional, one or other of the above methods will enable us to reduce the exponent of x until its value lies between 0 and 1, or 1, and we shall then be enabled to approximate to the required integral by series.

On Logarithmic Functions.

Let it be required to integrate

Xda log." *

where X is any algebraic function of x.

If n is a positive whole number we may integrate by the method of parts, regarding log." as constant in the first instance. We shall then have

and since fx dx is supposed to be known by the principles already establish

X

ed, we perceive that the integration of the proposed function is reduced to that of one whose form is the same, and in which the exponent of the logarithm is reduced by unity. The same process is applicable to this new function, and thus the integration will be completed step by step.

Adding the successive results obtained in this manner, we find,

æ dæ log." x = m +1 log."x_nlog."- ̧n(n—1)log."—2x
(m +1 (m + 1)2 + (m+1)3

But it n be integral and negative, we perceive that, as in the case of exponential functions, in performing the integration by parts of ƒ X log." x dx, we must in the first instance suppose X constant.