Resolution of Homework 5 on Calculus with Analytic Geometry II | MATH 141, Assignments of Analytical Geometry and Calculus

Download Resolution of Homework 5 on Calculus with Analytic Geometry II | MATH 141 and more Assignments Analytical Geometry and Calculus in PDF only on Docsity! Homework 5 Solutions Section 8.2: 1. (a) The correct form is: 2x (x+ 3)(3x+ 1) = A x+ 3 + B 3x+ 1 . (b) The correct form is: 1 x3 + 2x2 + x = 1 x(x+ 1)2 = A x + B x+ 1 + C (x+ 1)2 . 2. (a) The correct form is: x x2 + x− 2 = x (x+ 2)(x− 1) = A x+ 2 + B x− 1 . (b) The correct form is: x2 x2 + x+ 2 = Ax+B x2 + x+ 2 . 4. (a) The correct form is: x3 x2 + 4x+ 3 = x3 (x+ 3)(x+ 1) = A x+ 3 + B x+ 1 (b) The correct form is: 2x+ 1 (x+ 1)3(x2 + 4)2 = A x+ 1 + B (x+ 1)2 + C (x+ 1)3 + Dx+ E x2 + 4 + Fx+G (x2 + 4)2 . 7. Substitute u = x− 6 (du = dx, x = u+ 6) to get∫ x x− 6 dx = ∫ u+ 6 u du = ∫ du+6 ∫ du u = u+6 ln |u|+C = x−6+6 ln |x− 6|+C. 10. First we work out the partial fraction decomposition, 1 (t+ 4)(t− 1) = A t+ 4 + B t− 1 = A(t− 1) (t+ 4)(t− 1) + B(t+ 4) (t+ 4)(t− 1) = At−A+Bt+ 4B (t+ 4)(t− 1) = (A+B)t+ (−A+ 4B) (t+ 4)(t− 1) Now equating coefficients, we get A+B = 0 −A+ 4B = 1 1 2 which give us A = − 15 and B = 1 5 . So∫ dt (t+ 4)(t− 1) = −1 5 ∫ dt t+ 4 + 1 5 ∫ dt t− 1 = −1 5 ln |t+ 4|+ 1 5 ln |t− 1|+ C. 11. First we work out the partial fraction decomposition, 1 (x+ 1)(x− 1) = A x+ 1 + B x− 1 = A(x− 1) +B(x+ 1) (x+ 1)(x− 1) = Ax−A+Bx+B (x+ 1)(x− 1) = (A+B)x+ (−A+B) (x+ 1)(x− 1) . Now equating coefficients, we get A+B = 0 −A+B = 1 which gives us A = − 12 and B = 1 2 . So∫ dx x2 − 1 = −1 2 ∫ dx x+ 1 + 1 2 ∫ dx x− 1 = −1 2 ln |x+ 1|+ 1 2 ln |x− 1|+ C. Now adding the bounds of integration,∫ 3 2 dx x2 − 1 = ( − 1 2 ln |x+ 1|+ 1 2 ln |x− 1| ]3 2 = ( − 1 2 ln(4) + 1 2 ln(2) ) − ( − 1 2 ln(3) + 0 ) = 1 2 ( ln(2) + ln(3)− ln(4) ) 12. First we work out the partial fraction decomposition, x− 1 x2 + 3x+ 2 = x− 1 (x+ 1)(x+ 2) = A x+ 1 + B x+ 2 = A(x+ 2) +B(x+ 1) (x+ 1)(x+ 2) = Ax+ 2A+Bx+B x2 + 3x+ 2 = (A+B)x+ (2A+B) x2 + 3x+ 2