Explanation Given: ∫ d t ( t − 3 ) 2 ( t + 4 ) Key Concept and Formula Used: The concept of partial fractions is used. In partial fractions, the denominator is factorised as much as possible and then the given expression whose integral is to be found is written equal to the sum of reciprocals of each factor of the denominator with constant coefficients and then value of constants is found by equating the equation. Hence, we find the integral of the result equal to the given integral. Concept of substitution is used, and formula used ∫ x n d x = x n + 1 n + 1 + C • ∫ 1 x d x = log x + C log a − log b = log ( a b ) Calculation: ∫ d t ( t − 3 ) 2 ( t + 4 ) Now, let 1 ( t − 3 ) 2 ( t + 4 ) = A ( t − 3 ) + B ( t − 3 ) 2 + C ( t + 4 ) Taking LCM both the sides 1 ( t − 3 ) 2 ( t + 4 ) = A ( t + 4 ) ( t − 3 ) + B ( t + 4 ) + C ( t − 3 ) 2 ( t − 3 ) 2 ( t + 4 ) 1 = A ( t 2 − t − 12 ) + B ( t − 4 ) + C ( t 2 + 9 − 6 t ) 1 = t 2 ( A + C ) + t ( − A + B − 6 C ) + ( − 12 A − 4 B + 9 C ) Equating like terms we get A + C = 0 A = − C ........ ( 1 ) − A + B − 6 C = 0 F r o m ( 1 ) − ( − C ) + B − 6 C = 0 C + B − 6 C = 0 B − 5 C = 0 B = 5 C ...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8, Problem 35CRE

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The value of the given integral.

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Chapter 8, Problem 34CRE

Chapter 8, Problem 36CRE

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Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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The value of the given integral.

Solution Summary: The author explains the concept of partial tions, wherein the denominator is factorised and the value of constants is found by equating the equation.

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The value of the given integral.

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Chapter 8 Solutions