PostClass 7.4-solutions

# 4 16 x4 thus i x4 16 x4 dx consequently thus i 2 1 4

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: r comparing coeﬃcients gives A = −B , C = 0, 1 A= . 4 16 . x+4 Thus I= x−4+ 16 x+4 dx . Consequently, Thus I= 2 1 4 1 x 1 dx −2 x x +4 I= 1 1 ln(x) − ln(x2 + 4) 4 2 = = x2 1 ln 2 8 x +4 2 with C an arbitrary constant. 1 006 2 1 x2 − 4x + 16 ln |x + 4| + C 2 . Determine the integral Consequently, I= I= 5 1 ln 8 2 10.0 points dx . (x − 1)(x + 3) . 1. I = 005 2. I = 1 ln 4 x−1 x+3 3. I = 1 ln 2 2x + 2 (x − 1)(x + 3) 10.0 points Determine the integral I= 1 ln (|(x − 1)(x + 3)|) + C 2 x2 dx . x+4 1. I = x2 − 4x + 4 ln |x + 4| + C + C correct 4. I = ln (|(x − 1)(x + 3)|) + C +C eakin (ace669) – PostClass 7.4 – morales – (56430) 5. I = 1 ln 4 x+3 x−1 5. Area = 6 + +C Explanation: By partial fractions, 1 A B = + , (x − 1)(x + 3) x−1 x+3 4 77 sq.units ln 42 Explanation: The graph of f is an even function opening upwards, while the graph of g is a horizontal straight line. Thus the required area is similar to the shaded region in the ﬁgure below...
View Full Document

## This homework help was uploaded on 03/21/2014 for the course M 408S taught by Professor Stepp during the Spring '11 term at University of Texas at Austin.

Ask a homework question - tutors are online