and, therefore, we assume = B + (æ+1)2 C D E + (x+1) + (x-1)2 + (x−1); which, reduced to a common denominator, gives the equation 23+22+2=A(x2-1)2 + Bx (x-1)2+Cx (x+1) (x−1)2 + Dx (x+1)2 + Ex (x+1)2 (x—1). A Let x= 0; then 2 = 1; then 4 = 4 B.. B = 2 x= Substitute these values of A, B, D, in the above equation, and we have 23+x2+2=2(x2-1)2 — x (x − 1)2 +x (x+1)2 + Cx (x+1) (x − 1)2 + Ex (x+1)(x-1), or, 22(x-1)+ ↓ x (x2−1) = Cx (x+1)(x-1)2+Ex (x+1)2 (x−1) and, dividing both sides by x (x+1) (x−1), we have :. x = A (x2+x+1)+Bx (x−1)+C (x−1) =(A+B) x2+(A-B+C) x+A-C; hence, by the method of undetermined coefficients, we have = // 1 tan { log (x−1) — log √x2+x+1+√/3 tan IV. Let it be required to integrate du = Ax+B (x+a")" dx. -1 √3 2x+1 √3 A and u = dx 2 (n−1) (22 + a2) "—1 + B / (x2 + a2)* To obtain the integral of the differential dx (1) Differentiate this equation, and reduce to a common denominator; then, 1 = {(H+K) — 2 (n−1) H} x2+(H+K) a2 .. (H+K) a2 = 1, and 2 (n-1) H = H+K; a formula which, by successive operations, diminishes the exponent n, and ..x1+2x3+3x2+3= Ax+B+ (Cx+D) (x2+1)+H (a2+1)2..... (1) Let x2+1=0, or x2 = hence eq. (1) becomes by substituting for Ax+B its value — 2x+1 or (x2+1)(x2+2x+2)= (Cx+D) (x2+1)+H (x2+1)2 = Cx+D+H (x2+1). . Let x2+1 = 0; then, as before, we have 2x+1 = Cx+D; and this, substituted in (2), gives x2+1 = H (x2+1), or H = 1. (2) + (x2+1) (2+1)2 x2+1 + By formula (A) we have, making n = 3, and n = 2, 5 12 2x + 2 + C. 1 tan -1 2x+1 11 2 (3.) — log (x−1)+9 log (x−2) — log (x−3)+C. 2 (4.) — log (x+1)— ¦ log (x+2)+ 1 log (x−3)+C. (5.) — log (2-1) — 4 log (x+2)+ (8.) 2 + x+log (x−2)6(x+3)3 +C. (12.) ¦ log (x−1)— 2log (x+1)+ log (x2+1)+C= log c x+1 ON THE INTEGRATION OF IRRATIONAL FUNCTIONS.* The first class of integrals which we shall consider are comprehended under the general form The integrals belonging to this class are, for the most part, obtained by the method of parts. Let us first take those of the form * The limits and nature of the present work will not permit us to enter at any length upon a subject so extensive and intricate as the Integral Calculus. We shall therefore merely indicate the process by which some of the most useful integrals may be found, and refer the student who wishes to prosecute this subject to the masterly work of Lacroix. C c c 2 |