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1. Integrate by parts S In(2x + 3)dx Let u = In (2x + 3), dv = dx Then du = (2/(2x + 3)) dx
1. Integrate by parts S In(2x + 3)dx Let u = In (2x + 3), dv = dx Then du = (2/(2x + 3)) dx and v = x S In(2x + 3) dx = x In(2x + 3) - S( 2x 2x+3 dx 2x But S dx = S(1 - 3 dx = x - (3/2)In (2x + 3) + C1 2x+3. 2x+3. So f In(2x + 3) dx = x In(2x + 3) - x+ In(2x + 3) + C 2 marks for selecting u and dv 4 marks for completing the second step 4 mark for completing the problem Deduct 2 marks if C is missing
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Introduction To Health Care Management
Authors: Sharon B. Buchbinder, Nancy H. Shanks
3rd Edition
128408101X, 9781284081015
