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Unformatted text preview: (22) + A 2 (2 + 5) ⇒7 = 0 + 7 A 2 ⇒ A 2 =1 2. Set t =5 in Equation (2), we have59 = A 1 (52) + A 2 (5 + 5) ⇒14 =7 A 1 + 0 ⇒ A 1 = 2 Plug in the values of A 1 and A 2 back into Equation (1), we have t9 t 2 + 3 t10 = 2 t + 51 t2 Hence the integral is Z t9 t 2 + 3 t10 dt = Z ± 2 t + 51 t2 ² dt = 2 ln  t + 5  ln  t2  + C (Or if you like, you can further simplify the answer into ) = ln ³ ³ ³ ³ ( t + 5) 2 ( t2) ³ ³ ³ ³ + C...
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This note was uploaded on 09/22/2011 for the course MATH 2153 taught by Professor Staff during the Fall '08 term at Oklahoma State.
 Fall '08
 staff
 Calculus, Fractions

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