Homework8

# Homework8 - MA 115 Homework Solutions for Week 8 5.7#18 x 1...

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Unformatted text preview: MA 115 Homework Solutions for Week 8 5.7 #18 x- 1 x 2 + 3 x + 2 = A x + 1 + B x + 2 . Multiply both sides by ( x + 1)( x + 2) to get x- 1 = A ( x + 2) + B ( x + 1). Substituting- 2 for x gives- 3 =- B ⇔ B = 3. Substituting- 1 for x gives- 2 = A . Thus, integraldisplay 1 x- 1 x 2 + 3 x + 2 dx = integraldisplay 1 parenleftbigg- 2 x + 1 + 3 x + 2 parenrightbigg dx = [- 2 ln | x + 1 | + 3 ln | x + 2 | ] 1 = (- 2 ln2 + 3 ln3)- (- 2 ln 1 + 3 ln2) = 3 ln3- 5 ln2 [or ln 27 32 ] 5.7 #21 10 ( x- 1)( x 2 + 9) = A x- 1 + Bx + C x 2 + 9 . Multiply both sides by ( x- 1)( x 2 + 9) to get 10 = A ( x 2 + 9) + ( Bx + C )( x- 1) (*). Substituting 1 for x gives 10 = 10 A ⇔ A = 1. Substituting 0 for x gives 10 = 9 A- C ⇒ C = 9(1)- 10 =- 1. The coefficients of the x 2-terms in (*) must be equal, so 0 = A + B ⇒ B =- 1. Thus, integraldisplay 10 ( x- 1)( x 2 + 9) dx = integraldisplay parenleftbigg 1 x- 1 +- x- 1 x 2 + 9 parenrightbigg dx = integraldisplay parenleftbigg 1 x- 1- x x 2 + 9- 1 x 2 + 9 parenrightbigg dx = integraldisplay 1 x- 1 dx- integraldisplay x x 2 + 9 dx- integraldisplay 1 x 2 + 9 dx...
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Homework8 - MA 115 Homework Solutions for Week 8 5.7#18 x 1...

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