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63. J f 3^-m+21 65 J (*-2) • J( * + * + l ) * *-2χ +Ίχ+4 67, 68. (xJ 5x -1) (*+1) dx. • J 69 • J* + 64 * J (χ-iy (*+3) "■ J * +6* +25 * J 3* (x-l) +x-2 „ j * + 8 * + 16 J (*-l) (* + l) dx. 57, n—2— 3 A d / ■ *· 60. 2 3 2 2 4 (Λ;-1)(Λ: 2 +4Λ:+29) 4*3+9*2+4ΛΓ+1 άχ. *4+3*3+3Λ:2+Λ: άχ 2 άχ l 64. άχ J *3(Λ:-1)2(Λ:+1)· 66. 2 dx 2 3 άχ 4 άχ 72. 2 3 djc. 2 9JC -3JC -23JC +30JC 4 dx. 2 2 χ -2χ +5χ-$ 4 2 -11JC2 + 5 * + 4 70. 4 3 djc. : 2 (JC 2 +4JC+8) 3 ' 4 άχ. άχ. 2 74. 3 2 § 5. e. m. of the different numbers q.

7n - f d _ f *2 + l - * 2 * " J (x + l)» " J (*2 + l) n = / 2 x2 f n-l-J (Λ2 + 1)η d *' We integrate the second integral by parts. w d = 28 PROBLEMS AND METHODS IN ANALYSIS Then du = dx, Therefore, Γ x2 J (x2 + l)" X v=^ τττ-α—^—7. 2(/i-l)(x 2 + l) r i - 1 -x f dx 2 1+ ~ ( 2 « - 2 ) ( x + l ) ^ - J (2«-2)(x 2 + l ) n - 1 ? ι ν Tx +i I Γ dx A = —5—7 = arc tan x. Therefore, : dx -2 + l) 3 J (* + n_1 1 x 3 x 3 arc 7~2 2 4 (x2 + l) 2 +' "F" 8 (x + l)TV + ~F" 19. Integrate 22 ' tan x + ^· dx - 4 x + 13)2 ' Solution.

3 - 2 Χ - Χ 2 = (3+Χ)(1-Λ;)5 thus, — 3 < x < 1. Rewrite the function in the form A2 — X2. 3-lx-x2 = 4 - ( x + l) 2 . ) dw = y 4 arc sin y + — Uy/(4-u2) + C. Hence, J = 2 arc sin — (x + l ) + y 0 + 1) V(3-2x-x 2 ) + C. [Note that there may in fact be neater methods of solving such integrals by making use of a suitable trigonometric substitution. ] 9. Integrate f J x* 22 -+ 6x + 14 dx. V(*-22 +6x + ll) Solution. We write the numerator and denominator in terms of the appropriate perfect square. (x+3)a+5 V{(x+3) 2 + 2} J V{< f (x+3) 2 dx, Γ 5 J V{(x+3)2 + 2} aX " i "J V{(x+3)2 + 2 } a x ' 42 PROBLEMS AND METHODS IN ANALYSIS = l ( x + 3)V(x 2 + 6x + l l ) - l o g | x + 3 + V(^ 2 + 6x + l l ) | + 5 1 o g | x + 3+V(* 2 + 6 * + Hence, 11 )l+C· 7 = 1 (x + 3) J(x2 + 6x + U)+4 log | x + 3 + V(* 2 + + 6JC + 1 1 ) | + C .

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