...(x — o)~(x + o)(x — o)(x — a) Let us now make a? A , A' , B (x-aY(x + a) ~(xa,y+ (xa) +x + a' Reducing the terms of the second member to a common denominator, we have and developing, and comparing the coefficients of the like powers of x, we obtain the equations A'+B=l,...

...+bx 2 . x(a — x) (a + x) ax. Let us now make , x(a — x)(a + x) xa — xa + x' 19 =^ + -*- + .° reducing the terms of the second member to a common denominator, we have a 3 + bx 2 _Aa 2 —Aa?+ Sox + Ba?+ Pax — Ca? x(a — x)(a and, comparing the like powers of x (Alg....

...ax. x(a — x)(a + x) Let us now make a3 + Z^ _A B , C • — r • x(a — x)(a + x) xa — xa + x reducing the terms of the second member to a common denominator, we hare cf+la? _A(f— .4-rf- Bax±Bj~ + Cax— Cx2 x(a—x)(a + x)~ x(a — x) and, comparing the like...

...this case, take the example ax + c ax + c ax -f- c A JB Assume = — | -- - where A and B are unknown constants whose values are to be determined by the...second member to a common denominator, we have ax + c _ A(x + b) Ex _ Ax + Ab + Ex x2 + bx ~ x2 + bz x2 + bx~ z2 + bx Hence ax + c = Ax + Ab + Bx ; and since...

...this case, take the example ox+ c ax + c ax + c AB Assume = -- -- - where A and B are unknown 3? + bx xx + b constants whose values are to be determined...the second member to a common denominator, we have ax_+_e__A(x+K}_ Ex _Ax + Ab + Bx x2 + lx ~ x2 + bz~ + x2 + bx ~ x2 + bx Hence ax + c = Ax + Ab + -Bx;...

...fractions integrated separately. A _ + ? 4 _C ÍL. To v + a)" (x + a>"~1 (* + a;1—» a; -f- a bring the terms of the second member to a common denominator, we have to multiply В by x + а, С by (x + a)» - and A" by (x + a)"-1; hence the numerators, when added...

...~^ c it ~~ ' У = г ~ x(x + b) ' Assume — - 1— = — | -- r where A and В are unknown x* + ox xx + b constants whose values are to be determined...Reducing the terms of the second member to a common denomi nator, we have ax + с _ A(x + b) Bx _Ax 2 ~ * > ~ Bx + bx ~ x* + bx x> + bx ~ x¿ + bx Hence...

...+ b) ax -\- с А. В Assume — - т— = — | -- 7 where A and В are unknown x* + ox xx -\- о constants whose values are to be determined by the...Reducing the terms of the second member to a common denomi nator, we have ax + с _ A(x + b) Bx _ Ax + Ab + Bx x2 + bx ~ x2 + bx x1 + bx ~ x¿ + bx Hence...

...— + т н т , ж, 1 — x, and 1 + x being the unequal factors of x — x3 (115, 117)- Bringing the terms of the second member to a common denominator, we have x* —2 _ A — Ax' + Bx + Bx* + Cx — Cx* Щ J.3 "" " yj 1 _ x\l\ j. z\ • Hence x* — 2 = A +...

...ж, and 1 + x being ж — хл х 1-х l + х the unequal factors of x — x3 (1.15, 117). Bringing the terms of the second member to a common denominator, we have z' — a _ A — Ax* + Bx + Bx* + Cx — Cx* яс — х* ~ x(l — x)(l + x) Hence x* — 2 = A +...