I need help w/ figuring out how to do this problem, so a step-by-step answer would be wonderful. Do i do long division? Thanks!
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∫ (x^4- 3x³ - 4x² + 29x - 44) / (x² - 9) dx
= ∫ x² - 3x + 5 dx + ∫ (2x + 1) / (x² - 9) dx ...... by long division
... (2x + 1) / (x² - 9) = A / (x - 3) + B / (x + 3)
... 2x + 1 = A (x + 3) + B (x - 3)
... -5 = -6 B ... ⇒ B = 5/6 ....... let x = -3
... 7 = 6 A ... ⇒ A = 7/6 ....... let x = 3
= ∫ x² - 3x + 5 dx + 7/6 ∫ dx / (x - 3) + 5/6 ∫ dx / (x + 3)
= 1/6 ((2x³ - 9x² + 30x) + 7 ln|x - 3| + 5 ln|x + 3|) + C
Answer: 1/6 ((2x³ - 9x² + 30x) + 7 ln|x - 3| + 5 ln|x + 3|) + C
= ∫ x² - 3x + 5 dx + ∫ (2x + 1) / (x² - 9) dx ...... by long division
... (2x + 1) / (x² - 9) = A / (x - 3) + B / (x + 3)
... 2x + 1 = A (x + 3) + B (x - 3)
... -5 = -6 B ... ⇒ B = 5/6 ....... let x = -3
... 7 = 6 A ... ⇒ A = 7/6 ....... let x = 3
= ∫ x² - 3x + 5 dx + 7/6 ∫ dx / (x - 3) + 5/6 ∫ dx / (x + 3)
= 1/6 ((2x³ - 9x² + 30x) + 7 ln|x - 3| + 5 ln|x + 3|) + C
Answer: 1/6 ((2x³ - 9x² + 30x) + 7 ln|x - 3| + 5 ln|x + 3|) + C
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∫ (x^4-3x^3-4x^2+29x-44)/(x^2-9) dx
= [ x( 2x² - 9x + 30 ) + 7log( 3 - x ) + 5log(x + 3) ]/6 + C
= [ x( 2x² - 9x + 30 ) + 7log( 3 - x ) + 5log(x + 3) ]/6 + C