Question

Evaluate the following integrals, which have irreducible quadratic factors. 207. 2 -dx (x – 4) (x2 + 2x + 6) s . x2 208. -dx x3 – x2 + 4x - 4 209. x3 + 6x2 + 3x + 6 -dx x3 + 2x2 _ - 210. -dx (x - 1)(x2 + 2x + 2) Use the method of partial fractions to evaluate the following integrals. 3x + 4 211. -dx (x2 + 4) (3 – x) j so 212. 2 -dx (x + 2)? (2 – x)
208 and 212 please ❤️

Answer 1

Wel dx 22 22 x ² x ²4 x 4 da x² (26-1) 44(x-1) 22 ♡ (21-1) (42+4) let 2² A (x-1) (x²+4) 1 Batc (22+4) X-1 2 x2 A Cx+4) + (BX+C)(x-1) → x ² = A x² +40 + Bx²+ (x-Bx-c. 22 2 = x² (A+B) + x (C-B) + (4A-C) by comparing terms, A+B te C=B. 2 4 A } cHD O = 50 4 A فيستي A !

hence I. Sit la (2-0) 5 (x+4) 1 da 4. 5 Sk da x ²4.4 x²44 5 - I (x-1) + 1 x 1 In (x² + 4) + 2 x tan 2. 5 + c In (x²+4) + tan (3)/+C In (2-1) + I 2 da (012 .da (347)2 (2-0) (2+2)2 (2 B Jet А (2-2) (2+2) (x + 2)2 1 (6+2) 2 (7-2) A = 12 +27 2/2 = 2 s ay ( 3-0) |222 78= c=2212=-2 = c = - 1 / 2 = | 2 (د ) A B = 8

I Straz) star + stores + + 12 +972 1dx 1 = Ini In 18-2) + 1 3 1n (2+2) - 2 (7-2) + c) G 8 니 8 Ans