5.

Find and explain the errors (if any) in the following solution. We evaluate 1+dt by first factorizing the denominator as follows. Notice 1 + t' = (1 + 12)2 - 212 and factoring we have 1 + t = (1+12)2 -212 = (1+12 - V2t) (1+ +2 + v2t). Considering the concept of partial fractions we have At + B Ct + D +2 + V2+ + 1 +2 - V2t+ 1 (which implies 1 = (At + B)(12 - v2t + 1) + (Ct+ D)(+2 + v2t + 1)) where A, B. C and D are determined by solving the following system of equations 1 = B + D, 0 = A - V2B + C + V2D, 0 = -V2A + B + v2C+D, 0 = A +C. The solution to this system of equations is A = -C = ,,, B = D = ;. We can now write 1 aS 1 21/2 1+ 14 12 + 2t + 1 +2 - V2t+1 1 t+ v2 1 t - V2 2V/2 (12 + V2+ + 1) 2V/2 (t2 - V2t + 1) 1 t+ v2 1 t - V2 2V/2 ((t + V2)2 + !) 2V/2 ((t - V2)2 + 7)where in the last line we completed the square. It follows that + + V2 It ldt = -dt - t - V2 dt 2V/2 ( (t + 12) 2 + !) 2V/2 ((1 - 13)2 + ;) let w = t+ and let u = t - 2 2 w+ 2 dw - 2V/2 2V/2. 1 W 1 1 Tdu + 7 1 du + -dw - du 2V/2. w2+ 191-1 w/2 - 191- 2V2 72 + let p = w- + and let q = u- + 1 AVE . Lap + 1 . -du 2 2+ -dw - AVZ - do + - / 72 + IVE . - dp + 2 . (V2w)2+1 dw - dg + 7 1 -du. (VZu)2+ 1 We know the antiderivatives of these integrals and hence evaluating we have If todt 1 1 arctan(v2w) - In |ql + . arctan(v2u) 4V2 In |pl + 2V/2 4V/2 2V/2 arctan( v2w) - 1/7 = arctan(v2u) 4V/2 - In |w= + 5+ 2V/2 In lu + 5+ 2V/2 1 1 In t + + 4V2 2V/2 arctan ( v7 (1+ 12) 2 1 In + + 2V/2 arctan (V/2 (1 - V2 4V/2 2 1 In t+ + + arctan 4V/2 2V/2 2 In + 2V/2 arctan 4V (VE ( 1 - 4 2 ) In the last line, the absolute values have been removed because ( + 12 ) + ; and (1 - 12) + ; are positive for any value of t.Evaluate 1+13 it using a similar approach as in question 5 (but without the errors, if there were any).