PostClass 7.4-solutions

0 points and so evaluate the integral 1 i 0 1 i i 1 dx

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

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

Unformatted text preview: ’4 3 ln 4 u B = βˆ’C , 1 0 x2 x dx +1 1 dx , +1 10.0 points and so Evaluate the integral 1 I= 0 1. I = I= 1 dx . (x + 1)(x2 + 1) Ο€ 1 ln(2) + 4 2 correct 1 1 ln(x + 1) βˆ’ ln(x2 + 1) + tanβˆ’1 (x) 2 2 = Consequently, Ο€ 1 ln(8) βˆ’ 2. I = 2 2 3. I = 1 Ο€ ln(8) βˆ’ 4 2 1Ο€ βˆ’ ln(2) 4. I = 22 5. I = I= Explanation: Ο€ 1 ln(2) + 4 2 004 10.0 points Evaluate the integral 2 I= 1Ο€ βˆ’ ln(2) 42 Ο€ 1 ln(2) + 6. I = 2 2 (x + 1)2 1 ln 2 + 2 tanβˆ’1 (x) 4 x +1 1 1. I = 8 1 ln 4 5 2. I = 8 1 ln 16 5 1 dx . x3 + 4 x . 1 0 . 1 0 eakin (ace669) – PostClass 7.4 – morales – (56430) 3 3. I = 1 5 ln 4 2 2. I = x2 βˆ’ 4x + 16 ln |x + 4| + C 4. I = 8 1 ln 8 5 3. I = x2 + 4x + 16 ln |x + 4| + C 1 5 5. I = ln correct 8 2 4. I = x2 + 4x + 16 ln |x + 4| + C 2 1 5 ln 16 2 5. I = x2 + 4x + 4 ln |x + 4| + C 2 6. I = x2 βˆ’ 4x + 16 ln |x + 4| + C correct 2 6. I = Explanation: By partial fractions, 1 A Bx + C = +2 . x3 + 4 x x x +4 Explanation: After division we see that To determine A, B, and C multiply through by x3 + 4x: for then x2 βˆ’ 16 16 x2 = + x+4 x+4 x+4 1 = A(x2 + 4) + x(Bx + C ) = xβˆ’4+ = (A + B )x2 + Cx + 4A , which afte...
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.

Ask a homework question - tutors are online