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I. Multiple Choice - Integration Technigues El 0 For each integral, select the most efficient integration technique or basic form that corresponds to that integral. 0 Shade in or circle the number next to that option. 0 In each case, it is understood that u-substitution may be used in conjunction with the given method. The option \"ts-substitution only" means that none of the other listed techniques are to be used. 0 Finally, provide any information requested for a technique by filling in the blank(s) with the information specific to your integral. 0 DO MEVALUATE THESE INTEGRALS!!! [1] Natural Logarithm form [2] Partial Fractions: Number of fractions: I 95" cosx dx [3] Inverse Trigonometric form: function [4] Trigonometric Substitution: x = [5] Integration By Parts: Let u = and dv = du = v = [6] Trigonometric Integral: Trig. Identity: [7] usubstitution oniy: Let u = then du = 2x [1] Natural Logarithm form [de [2] Partial Fractions: Number of fractions: [3] Inverse Trigonometric form: function [4] Trigonometric Substitution: x = [5] Integration By Parts: Let u = and dv = du = v = [6] Trigonometric Integral: Trig. Identity: [7] usubstitution oniy: Let u = then du = 6x [1] Natural Logarithm form I 3x2 _ 5 dx [2] Partial Fractions: Number of fractions: [3] Inverse Trigonometric form: function [4] Trigonometric Substitution: x = [5] Integration By Parts: Let u = and (it? = du = v = [6] Trigonometric Integral: Trig. Identity: [7] usubstitution only: Let u = then du = 4x [1] Natural Logarithm form I x2 _ 16 dx [2] Partial Fractions: Number of fractions: [3] Inverse Trigonometric form: function [4] Trigonometric Substitution: x = [5] Integration By Parts: Let u = and (ii? = du = v = [6] Trigonometric Integral: Trig. Identity: [7] usubstitution only: Let u = then du = x [1] Natural Logarithm form m dx [2] Partial Fractions: Number of fractions: [3] Inverse Trigonometric form: function [4] Trigonometric Substitution: x = [5] Integration By Parts: Let u = and dv = du = v = [6] Trigonometric Integral: Trig. Identity: [7] usubstitution only: Let u = then du = f sinzx dx [1] Natural Logarithm form [2] Partial Fractions: Number of fractions: [3] Inverse Trigonometric form: function [4] Trigonometric Substitution: x = [5] Integration Bv Parts: Let u = du = [6] Trigonometric Integral: Trig. Identity: [7] usubstitution only: Let u = [1] Natural Logarithm form [2] Partial Fractions: Number of fractions: [3] Inverse Trigonometric form: function [4] Trigonometric Substitution: x = [5] Integration Bv Parts: Let u = du = [6] Trigonometric Integral: Trig. Identity: [7] usubstitution only: Let u = f L [1] Natural Logarithm form V2 25x2 [2] Partial Fractions: Number of fractions: [3] Inverse Trigonometric form: function [4] Trigonometric Substitution: x = [5] Integration By Parts: Let u = and dv = du = v = [6] Trigonometric Integral: Trig. Identity: [7] usubstitution only: Let u = then du =