7.7.95-eg-1 - u = x-2 3 , then du = 1 3 dx . A substitution...

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Example Evaluate Z dx x 2 - 4 x + 13 . We start by completing the square in the denominator, x 2 - 4 x + 13 = x 2 - 4 x + ± - 4 2 ² 2 - ± - 4 2 ² 2 + 13 = ( x - 2) 2 + 9 . We can rewrite the integral as Z dx x 2 - 4 x + 13 = Z dx ( x - 2) 2 + 9 = 1 9 Z dx ( x - 2) 2 9 + 1 = 1 9 Z dx ³ x - 2 3 ´ 2 + 1 . Now let
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Unformatted text preview: u = x-2 3 , then du = 1 3 dx . A substitution yields Z dx x 2-4 x + 13 = 1 3 Z du u 2 + 1 = 1 3 tan-1 u + C = 1 3 tan-1 x-2 3 + C....
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