MATH300 Final: MATH 300 UofA Solution Final Exam
Document Summary
Calculators and other electronic devices are neither allowed nor required for this test. = z dx x2 + 2x 3 (x 1)(x + 3) We will solve this problem using partial fractions: z dx x2 + 2x 3. A(x + 3) + b(x 1) (x 1)(x + 3) From rst equation a = b, adding both equations, 4a = 1. Therefore, a = 1 (we can also solve it by plugging x = 3 and x = 1 in a(x + 3) + b(x 1) = 1. ) Z ln|x + 3| + c ln|x 1| . 4(x 1) x2 + 2x 3 dx: z x cos(x)dx. Solution: we solve this problem using integration by parts. Let u = x and dv = cos(x). Then du = dx and a possible v = sin(x). Z x cos(x)dx = x sin(x) z sin(x)dx = x sin(x) + cos(x) + c: z ln(2x) x dx.
