The process is precisely analogous to that employed in the last case

2. Let

du =

The preceding integral may be found in a manner somewhat different

ON EXPONENTIAL FUNCTIONS.

1o. If X = ƒ (a*), then the function Xdz, if we make aa = u wil become

2o. Differentiating Xe, we have e* 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 examplo

In like manner, if we make 1 + x = z, we shall find

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

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

N x2-1 n(n−1)r-2 1.2.3..n
+
log.alog.a log. a

u

=a* { log.a

...

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 z 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

dx

Integrating in the same manner, we shall finally have

24-1