(19) S dx bx3) = 1 1 √ = (a + ba'y' = 3a (a + bx) — 3as log (a+ba"). x dx 1 3a2 1 2 x2 1 dx (21) Si d = log(+)+ & tan- a. Integrals of the form fdx x" (a + ba")" can be rationalized, + is an integer, by assuming a + bx" = ≈1, and, is an integer, by assuming a + bx” = x” z1. x dx (a + bx) x dx (x − 1)3 dx 1 1 1)3 + 3 (x − 1)2 + a} . 5 (3) Ja (a+ba); a) log ((a+b)) - all. x = (6) x3 − 1)1 (7) √(a + bx) = 2 J2a + bx b2 [(a + bx) 3) 7 4 5 = bo (a + bx-2). 7 2 3 (a + bx)2 + 6a (a + bx) - a2 3b3 (a + bx) # 3 x dx 3 bx (10) (a+ba); " (a+ba); (a+b2 - 9). x2 da (11) Sa+ = b2 5 2 7 (13) fdæ a3 (a + x)2 = 3 (a + x)3· 14 2 3a (a + x)2 3a2 (a + x) (a + bx2)3. (1 + x2)3. 6 2a + bx2 b2 (a + bx2) 3° x (x2 - 3) 2 (1-2) 8 5x3 5.3.x + 3 sin-1x. 2 dx x (2x2 + 3) = (1 + x2)2 3 (1 + a2)! · 3a (a + bx2) If an integral be a function of several fractional powers of a, it may be rationalized by assuming x = x', r being equal to the product of the denominators of the indices. When the integral involves also fractional powers of binomials, such as a + ba, it can be rationalized by assuming a + bx = ", r being the product of the denominators of the indices. If the binomials be of the form a+ba", they may be reduced to the preceding form by assuming a” = y. dx a e c ae>bc (90) (e+ex) (a+bx)) = sin"'z (br), ae: + aex c (a + bx) = 2 (c + ex3) 3 (31) Ja (a+b) na log x bx") 3 dx x2-1 = 1 (a - bax2n); nb if ae = bc. (a + bx")} — a3 ae <bc If the function to be integrated involve (a + bx + cx2)} it may be reduced to the preceding forms, as Various functions can be rationalized by assumptions for which no general rule can be given: familiarity with the transformations to which different substitutions lead is the best way of acquiring a knowledge of the most convenient assumption in particular cases. |
