Now here is an apparent inconsistency; for it would appear

must therefore have recourse to the accurate process of definite integration; and let us take both integrals between the same limits; say, between the limits O and x; then by (11), Art. 5,

which is a correct result; (61) and (62) therefore are not simultaneously true in the forms of indefinite integrals.

36.] The relations existing between some of these last integrals deserve consideration; for taking the definite integral of (69) between the limits a and 0, we have

x dx

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

0 (a2 + x2) 3

a

but the left-hand member may be put under the following form, whereby its integral is determined by equation (61): viz.

and the identity of (73) and (74) may thus be proved:

replacing sin (z√−1) in terms of its exponential value as given by equation (30), Art. 61, of Vol. I;

which shews that the two results though different in form are

identical in value. Similarly might

x

dx
a2 + x2

be put in the

1

Article 15, and the result shewn to be equal to -tan-1

a

whereby the integral is reduced to one or other of the forms

39.] The integration of (a2 — x2)3 dx, and of (a2 +x2)13 dx.

In these cases it is convenient to employ the method of integration by parts given by the formula,

Then, adding and subtracting a2 in the numerator of the last quantity, and writing the fraction in two parts, and cancelling in the second part (a2 —x2), which occurs in both numerator and denominator, we have

..

(a2 — x2)1

dx

= 2 (a2—22)1 + a2 sin−1′′ — ƒ (a2—22)1 dx ;

x —

α

a2

+ = sin-14

f(a2—22)dx = ~(a2 = 22)) + 2a sin

To determine [(a2+22)1 dæ.

Let u = (a2+x2)3,

x dx

2

2

dvdx;

(79)

du =

Then, following a process similar to that of the last integral,