8.1.47-eg-1 - Note that x + 1 x-1 = x + (-1 + 1) + 1 x-1 =...

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Example Evaluate Z x + 1 x - 1 dx . There are two ways to approach this problem. 1. Use long division. The integrand is an improper fraction, so using long division Z x + 1 x - 1 dx = Z 1 + 2 x - 1 dx. Make the substitution u = x - 1 ( du = dx ) into the second term to obtain Z x + 1 x - 1 dx = Z dx + 2 Z du u = x + 2 ln | x - 1 | + C. 2. Add and subtract one in the numerator, then split the integrand.
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Unformatted text preview: Note that x + 1 x-1 = x + (-1 + 1) + 1 x-1 = ( x-1) + 2 x-1 = x-1 x-1 + 2 x-1 = 1 + 2 x-1 . The rest of the problem is carried out the same as above....
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

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