5. Find and explain the errors (if any) in the following solution. We evaluate / 1 + :4 t2)2 2t2 and factoring we have (if by rst factorizing the denominator as follows. Notice 1 + t4 = (1 + 1 +t4 = (1 +t2)2 21:2 = (1 +t2 x/itxl +1:2 + x/it). Considering the concept of partial fractions we have 1 _ At+B + Ct+D 1+t4_t2+~/t+1 tgx/Et+1 (which implies 1 = (At + B)(i2 t + 1) + (Ct + D)(t2 + t + 1)) where A, B, C and D are determined by solving the following system of equations 1 = B + D, 0=A\\/B+C+\\/D, 0=\\/A+B+\\/C+D, 0 = A + C. The solution to this system of equations is A = C = 21W' B = D = i. We can now write 1-:t4 as 1 t + 31$: + % 1+t4_t2+\\/t+1 t2\\/t+1 _ 1 t+\\/ 1 t\\/ 2\\/(t2+\\/t+1)2\\/(32\\/t+1) 1 t+\\/ 1 t = 2~/((t+ sw) _ mat sm %) where in the last line we completed the square. It follows that t+ v2 t - V2 It tidt = dt dt 2V/2 ( (t + +2 ) 2 + 5) 2V/2 ( (t - 12)2 + 2 ) let w = t+ . V2 and let u = t V2 2 2 w+ V2 V2 2 dw - u - 2 du 2V2 w2 + 7 2V/2 . u2 + 1 w 1 2V/2 w2 + Idw+ / w2 + 2 dw - NIL let p = w' + - and let q = u- + 2 1 1 AV2J - ap + 1 / w 2 + ! dw 4V2 (V2w)2 + 1 dw - 4V/2 - da+ 2 / (V2u ) 2 + 1 -du. We know the antiderivatives of these integrals and hence evaluating we have 1 + ta dt 1 - In |pl + arctan(V2w) - -Ing+ 4V2 2V/2 4V/2 2V/2 arctan(V2u) 1 In / w2 + = 1 + 1 arctan(V2u) 4V2 2V2 arctan(V2w) - 1 In|u2+ 21 + 4V2 2V/2 2 1 In t + V2 2 + 2 + arctan 4V/2 2V/2 ( Vz ( + + 4 2 ) 1 In t - V2 + + 4V2 NI 2V/2 arctan ( v2 (t - 12 ) 2 1 1 In t+ + + arctan V2 4V2 2V/2 1 1 In + + 4V2 2V2 arctan ( v2 (t - 12 ) ) In the last line, the absolute values have been removed because (t + v2 ) + ; and (t - 2) 2 + 3 are positive for any value of t.1 6. Evaluate 1 + +3 at using a similar approach as in question 5 (but without the errors, if there were any)