1880.] History of Planet and Single Satellite. 263 



Again b> or <jk, 



as (\-3- + 7t)[l- > or <SJi, 



(XI + h) 7 



as — — ~i — v2h— Xf> or <2h— XL 



Since the left-hand side is negative and the right positive, the left is 

 less than the right, and therefore b is less than 

 If, therefore, h be greater than 4/3 3 , we may write 



where a— f ft, fft— b are positive, and where a is negative. 



We now turn to the other case and suppose ft less than 4/3 1 . All 

 the roots of the biquadratic are now imaginary, and we may put 



x*-hx* + l = [(x-*y+p*]\_(x- 1 y + &]. 



If a be taken as — J(Xt — ft), then 7 is J(Xf + ft). 



Then it only remains to prove that 7 is greater than £ ft. 



Now 7> or <fft, 



as XI- > or < 2h, 



as X 3 > or < 4ft 2 =4(X 3 — 4X), 



as 16 > or < 3X 2 , 



as 4/3* > or < X, 



but it] has been already shown that in this case, X is less than 4/3*, 

 wherefore 7 is greater than jft. 



We may now proceed to the required integrations. 

 First case where ft is greater than 4/3 1 . 



Let a3 4 -/^ 3 + l = («-a)(«3-b)[(a;- a ) 2 + ^ 2 ], 



so that the roots are a, b, « + /3f. 



Also let a be the root which is greater than jft, b that which is less, 

 and let 



a=a 1 + fft, b=fft— b l5 x=^h—a. l . 



To find the expression for i we have to integrate . 



r & x*-hx* + l 



Let /(aj) = (aj-a)f («), and let x 2 /f(x)=AI(x— &)+<]> (x)lyjr(x) . 



Then - o3 2 («-a)=4/(«) + (^-a) 2 0(a;). 



Hence A=a?/f(ti). * 



If, therefore, /(oj)=aj*-feu5 + l, A=l/(4a-3ft)=l/4a 1 . 



t 2 



