Homework8

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 04/07/2008 for the course MA 115 taught by Professor Mahalanobis during the Fall '08 term at Stevens.

Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online