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9x2 – 8x + 20 x²+ 4x EXAMPLE 5 Evaluate dx. SOLUTION Since x3 + 4x = x(x² + 4) can't be factored further, we write Вх + C 9x? – 8x + 20 x(x² + 4) Multiplying by x(x² + 4), we have 9x2 – 8x + 20 = A(x² + 4) + + Cx + 4A. Equating coefficients, we obtain A + B = 9 C = -8 4A = 20. , В 3 and C = -8 and so Thus A = 9x² – 8x +20 3 + 4x X - 8 dx. dx = x² + 4 In order to integrate the second term we split it into two parts: /- 8 х — 8 dx x² +.4 dx. x + 4 dx = x2 + 4 We make the substitution u = x + 4 in the first of these integrals so that du = 2x dx. We evaluate t integral by means of this formula with a = 2. (Remember to use absolute values where appropriate.) %D %3D 8 9x² – /- dx x2 + 4 9x2 - 8x + 20 dx dx + - хр x(x² + 4) + K

9x2 – 8x + 20 x²+ 4x EXAMPLE 5 Evaluate dx. SOLUTION Since x3 + 4x = x(x² + 4) can't be factored further, we write Вх + C 9x? – 8x + 20 x(x² + 4) Multiplying by x(x² + 4), we have 9x2 – 8x + 20 = A(x² + 4) + + Cx + 4A. Equating coefficients, we obtain A + B = 9 C = -8 4A = 20. , В 3 and C = -8 and so Thus A = 9x² – 8x +20 3 + 4x X - 8 dx. dx = x² + 4 In order to integrate the second term we split it into two parts: /- 8 х — 8 dx x² +.4 dx. x + 4 dx = x2 + 4 We make the substitution u = x + 4 in the first of these integrals so that du = 2x dx. We evaluate t integral by means of this formula with a = 2. (Remember to use absolute values where appropriate.) %D %3D 8 9x² – /- dx x2 + 4 9x2 - 8x + 20 dx dx + - хр x(x² + 4) + K

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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9x2 – 8x + 20 x²+ 4x EXAMPLE 5 Evaluate dx. SOLUTION Since x3 + 4x = x(x² + 4) can't be factored further, we write Вх + C 9x? – 8x + 20 x(x² + 4) Multiplying by x(x² + 4), we have 9x2 – 8x + 20 = A(x² + 4) + + Cx + 4A. Equating coefficients, we obtain A + B = 9 C = -8 4A = 20. , В 3 and C = -8 and so Thus A = 9x² – 8x +20 3 + 4x X - 8 dx. dx = x² + 4 In order to integrate the second term we split it into two parts: /- 8 х — 8 dx x² +.4 dx. x + 4 dx = x2 + 4 We make the substitution u = x + 4 in the first of these integrals so that du = 2x dx. We evaluate t integral by means of this formula with a = 2. (Remember to use absolute values where appropriate.) %D %3D 8 9x² – /- dx x2 + 4 9x2 - 8x + 20 dx dx + - хр x(x² + 4) + K
Transcribed Image Text:9x2 – 8x + 20 x²+ 4x EXAMPLE 5 Evaluate dx. SOLUTION Since x3 + 4x = x(x² + 4) can't be factored further, we write Вх + C 9x? – 8x + 20 x(x² + 4) Multiplying by x(x² + 4), we have 9x2 – 8x + 20 = A(x² + 4) + + Cx + 4A. Equating coefficients, we obtain A + B = 9 C = -8 4A = 20. , В 3 and C = -8 and so Thus A = 9x² – 8x +20 3 + 4x X - 8 dx. dx = x² + 4 In order to integrate the second term we split it into two parts: /- 8 х — 8 dx x² +.4 dx. x + 4 dx = x2 + 4 We make the substitution u = x + 4 in the first of these integrals so that du = 2x dx. We evaluate t integral by means of this formula with a = 2. (Remember to use absolute values where appropriate.) %D %3D 8 9x² – /- dx x2 + 4 9x2 - 8x + 20 dx dx + - хр x(x² + 4) + K
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