Explanation Given: ∫ x − 2 tan − 1 x d x Key Concept and Formula Used: By parts ∫ u . v d x = u ∫ v d x − ∫ ( d u d x ∫ v d x ) d x + C where u is first function and v is second function. To decide first and second function we use ILATE where I=Inverse function, L=Logarithmic function, A=Algebraic function, T=Trigonometric function, and E=Exponential function 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: • d d x tan − 1 x = 1 1 + x 2 • ∫ x n d x = x n + 1 n + 1 + C Calculation: ∫ x − 2 tan − 1 x d x Using by parts ∫ u . v d x = u ∫ v d x − ∫ ( d u d x ∫ v d x ) d x + C Here u = tan − 1 x & v = x − 2 ∫ x − 2 tan − 1 x d x = tan − 1 x ∫ x − 2 d x − ∫ ( d d x tan − 1 x ∫ x − 2 d x ) d x = tan − 1 x ⋅ x − 2 + 1 − 2 + 1 − ∫ 1 1 + x 2 ⋅ x − 2 + 1 − 2 + 1 d x + C (Using d d x tan − 1 x = 1 1 + x 2 & ∫ x n d x = x n + 1 n + 1 + C ) = tan − 1 x ⋅ x − 1 − 1 − ∫ 1 1 + x 2 ⋅ x − 1 − 1 d x + C = − ...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8, Problem 52CRE

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The value of the integral.

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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 integral.

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

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The value of the integral.

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