MATH300 Final: MATH 300 UofA Final Exam
Document Summary
Calculators and other electronic devices are neither allowed nor required for this test. We will solve this problem using partial fractions: z dx x2 + x 2. A(x + 2) + b(x 1) (x 1)(x + 2) From rst equation a = b, adding both equations, 3a = 1. Therefore, a = 1 (we can also solve it by plugging x = 2 and x = 1 in a(x + 2) + b(x 1) = 1. ) Z ln|x + 2| + c ln|x 1| . 3(x 1) dx x2 + x 2: z x sin(x)dx. Solution: we solve this problem using integration by parts. Let u = x and dv = sin(x). Then du = dx and a possible v = cos(x). Z x sin(x)dx = x( cos(x)) z ( cos(x))dx = x cos(x) +z cos(x)dx = x cos(x) + sin(x) + c: z ln(3x) x dx. Solution: we solve this problem by using substitution u = ln(3x).
