100. We have already (Art. 98) recommended the student to recollect the modes of investigating the different formulas of reduction which we have obtained, and to apply them to particular cases, instead of substituting in the general formulas the particular values of the indices and coefficients. We shall here give one or two instances of integration as patterns to be followed by the student in other cases. Commencing with the elementary forms, where m = 0 and m = 1, = x2√√(x2±a2) — 2u2 a2 Transposing2u, and dividing by 3, gives x dx = x2√ (x® ± a3) − 2 '(x2± (x2 + a2) x dx √(x2 ± a3)* In this manner we may ascend from one integral to another, until the proposed integral be obtained. 2 To integrate du = xTM (a2 + x2) ✈ dx. Let m = 2; then du = u = ƒ x.x dx (a* + x°) ↓ (a2 = x2 (a2 + x2) § dx, and, integrating by parts, = a2 3 3 Integrals of this class may be obtained in the following manner: thus In a similar manner, the integrals of various other differentials may be found; and if the power of x be a factor of the denominator, the integral may be readily found by assuming y 1 = INTEGRATION OF TRANSCENDENTAL FUNCTIONS. 101. Transcendental functions are such as have differential coefficients involving logarithmic, exponential, or circular functions, and we shall now advert shortly to these in their order. I. Logarithmic Functions. Let it be required to integrate du = x (log x)" dx. f(log x)". x dx = m+1 (log x)" n fx (log x)-1dx, which is the formula of reduction, the final integral being fTM dx= (log x)* + 32o log x — x dx (log x)** = 4 3x2 8 +1. (log x)" d (log x) m+ 1 + N x dx (log x)-1 xdx which is the formula of reduction, the final integral being log x which must be found by diminishing m successively. To reduce this to a simpler form, let +1 z, then (m + 1) log x = log z, and 102. Let it be required to integrate du = x" a dx, where m is positive. Integrating by parts, we have By the successive application of this formula, the index of x will be continually diminished. In a similar manner we get a* dx When m = 1, the formula evidently fails, since x" 0, and there cannot be found except by expanding a in a series. 1+ log a.x + (log a)2 + (log a)3. 1.2 The formula may be integrated by series in a similar manner; dz log z for if log z = x, or z = e, then we have log + log z+ 1 (log 2)2 2 1.2 where log z denotes the logarithm of log z. EXAMPLES. 1.2 3 1.2.3 1 (log 2)3 1. ƒ z (log x)' d z = { (log z) — log 2 + 1 }, fx x = = { (log x)" — + log x + +}. = = 4 x2 a* log a απ log a = = a* Xx {x {x хив e* log al 3x2 log a + log a far dz e e* dx x log x = a Seda + log (1 − x). 2ear x log x eax III. Circular Functions. 103. If the differential contains a variable arc, as tan - 'x, it may generally be made to disappear in the differential derived in the method of integration by parts. Thus |