Question

3. Calculate the following definite integral to the third-degree polynomial of k. 1(k)= 1 - kt dt, k > 0 Jo VI-E

Answer 1

Problem: 11- kt V 1-t Rewrite/simplify: 1 - kt dt 1 -t V1-t kv1-7 Substitute u =- di = dt = V1 - kt 1 2V1 - tv1 - ht (steps) 2(1 – kt) > ht 2(1-kt) ] —du: 2/1-1V1-kt = 2 (8 – 1./ (Ne? - 1,ja du = 2 (k - J (k ... or choose an alternative: 1-t Substitute 1- kt Substitute 1- kt 1-t

Substitute 1 – kt substitute1-t Now solving: (tz2 -1)2 du Factor the denominator: Perform partial fraction decomposition: - (= +ས) (ཝ ཙ ༩) སིང་ - (1( །ཡཟླ (གཞས་ཚོད་)མ V ད du Apply linearity: ༼ 1 1 1 - du + 4 །（ 4 = du + ， sg + ve Jku - vk

Now solving: f 1 ku + vã du Substitute v = ku + k = k (steps) + du = dv: du = 1/dv KJ Now solving: A do This is a standard integral: = In() Plug in solved integrals: By S5 dv In(v

Undo substitution t = ka+VL: In(ku+ V) Now solving: 11 du J (ku + vay?" Substitute v = ku + k +7 = k (steps) du = dv: = #Sudv Now solving: S do Apply power rule: 20+1 V = with n = n+1" -2

Plug in solved integrals: Ai Sadu kv Undo substitution v = ku + V: 1 k (ku+ V) Now solving: Substitute v = ku - vk = k (sters) → du = = dv: Adv kJ V

Use previous result: In(v) Undo substitution v = ku – Vk: In(ku – V) Now solving: 1 I Thou - Vojens 2 du Substitute v = ku - V =k (steps) du =- du: = du Use previous result: 1 Undo substitution v = ku – Vk:

1 k (ku – vā) Plug in solved integrals: kr 1 k 1 - dut A 4 Jku - vk (ku - ſk) ) 2 4 ) ku + V 41 (ku+V+) In(ku + Vk) In(ku – V) 452 45k 1 4 (ku + V) 4 (ku - VE) Plug in solved integrals: (k – 1) In(ku+V) 2vā 2 (k – 1). I Qu2 - 2ja du (k – 1) In(ku - VT k-1 k-1 25 2(ku + VE) 2(ku – VE)

Vl-t Undo substitution u = - ✓1 - kt. kv1- +VR) (k – 1) in (Ime - VK (k – 1) ln( 2 k-1 k-1 11-7 k 2k 2V 2(x* + v) 24 / kv1-t Il-Kt / kv1-t Wi-kt - va) The problem is solved. Apply the absolute value function to arguments of logarithm functions in order to extend the antiderivative's domain: MH a (k – 1) In ( 2V + v) (k – 1) In (KFT-VE -- k - 1 k-1 - +C = 2 k 2 (*+ V) 26 2 kv1-7 Il-kt kv1-t Il-kt k Rewrite/simplify: (k – 1) (In((VkV1 – kt + kV1 – t) ) – In ( l 2vã ivky 1 – kt – kv1 – t))) -V1- tv1 - kt+C

DEFINITE INTEGRAL: f(t) dt = (k – 1) Vk arcsin (4+1) + 2ik пk — п - + 2k 4k Simplify/rewrite: i (12k – 2) arcsin (4+1) + k + 4i/k – ) 417