Explanation Result used: Repeated linear factor : A factor ( x − r ) m with m > 1 in the denominator requires the partial fractions A 1 ( x − r ) + A 2 ( x − r ) 2 + ... + A m ( x − r ) m . Repeated irreducible quadratic factor : An irreducible factor ( a x 2 + b x + c ) m with m > 1 in the denominator requires the partial fractions A 1 x + B 1 a x 2 + b x + c + A 2 x + B 2 ( a x 2 + b x + c ) 2 + ... + A m x + B m ( a x 2 + b x + c ) m . Calculation: Obtain the value of ∫ 2 x 3 + 5 x 2 + 13 x + 9 x 2 ( x 2 + 4 x + 9 ) d x as follows. Frist rewrite the given fraction 2 x 3 + 5 x 2 + 13 x + 9 x 2 ( x 2 + 4 x + 9 ) as follows. 2 x 3 + 5 x 2 + 13 x + 9 x 2 ( x 2 + 4 x + 9 ) = x 3 + x 3 + 4 x 2 + x 2 + 9 x + 4 x + 9 x 2 ( x 2 + 4 x + 9 ) = x 3 + x 3 + 4 x 2 + 9 x + x 2 + 4 x + 9 x 2 ( x 2 + 4 x + 9 ) = x 3 + x ( x 2 + 4 x + 9 ) + ( x 2 + 4 x + 9 ) x 2 ( x 2 + 4 x + 9 ) = x 3 x 2 ( x 2 + 4 x + 9 ) + x ( x 2 + 4 x + 9 ) x 2 ( x 2 + 4 x + 9 ) + x 2 + 4 x + 9 x 2 ( x 2 + 4 x + 9 ) = x x 2 + 4 x + 9 + 1 x + 1 x 2 Now integrate the above expression as follows. ∫ 2 x 3 + 5 x 2 + 13 x + 9 x 2 ( x 2 + 4 x + 9 ) d x = ∫ x x 2 + 4 x + 9 d x + ∫ 1 x d x + ∫ 1 x 2 d x = I 1 + ln | x | + ∫ x − 2 d x + C [ ∵ ∫ x x 2 + 4 x + 9 d x = I 1 ] = I 1 + ln | x | + x − 1 − 1 + C = I 1 + ln | x | − 1 x + C Now obtain the value of I 1 as follows

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

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ISBN: 9780134996103

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

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Chapter 8, Problem 45RE

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To evaluate: The integral ∫2x3+5x2+13x+9x2(x2+4x+9)dx.

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Chapter 8, Problem 44RE

Chapter 8, Problem 46RE

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Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

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The integral ∫ 2 x 3 + 5 x 2 + 13 x + 9 x 2 ( x 2 + 4 x + 9 ) d x .

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To evaluate: The integral ∫2x3+5x2+13x+9x2(x2+4x+9)dx.

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Chapter 8 Solutions