Exam_solutions_3_

# Exam_solutions_3_ - MATH 141 3rd Examination—Solution Key...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 141, 3rd Examination—Solution Key Prof. Jonathan Rosenberg Friday, October 26, 2007 1. (25 points) Compute integraldisplay 5 x 2 + 2 x- 1 ( x 2 + 1)( x- 1) dx. Solution. Use partial fractions. 5 x 2 + 2 x- 1 ( x 2 + 1)( x- 1) = Ax + B x 2 + 1 + C x- 1 . Cross-multiplying, 5 x 2 + 2 x- 1 = ( Ax + B )( x- 1) + C ( x 2 + 1) . There are many correct ways to solve for A,B,C : expand both sides and equate coefficients of powers of x , substitute specific values for x , or some combination of both. Here is one method. Let x = 1 ; that gives 2 C = 6 or C = 3 . Equate coefficients of x 2 ; that gives 5 = A + C , or A = 5- C = 2 . Let x = 0 ; that gives- 1 =- B + C , so B = C + 1 = 4 . Upshot of all of this: 5 x 2 + 2 x- 1 ( x 2 + 1)( x- 1) = 2 x + 4 x 2 + 1 + 3 x- 1 . Integrating term-by-term gives integraldisplay 5 x 2 + 2 x- 1 ( x 2 + 1)( x- 1) dx = ln( x 2 + 1) + 4 tan- 1 x + 3 ln | x- 1 | + C. 2. (25 points) Compute integraldisplay 1 √ 4 + x 2 dx....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

Exam_solutions_3_ - MATH 141 3rd Examination—Solution Key...

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

View Full Document
Ask a homework question - tutors are online