# 8.3.9-eg - A 1 + A 2 ) x + (-2 A 1 ) . Since the coecients...

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Example Evaluate: Z dx x 2 - 2 x . Solution: To apply the method of partial fractions to the integrand, ﬁrst factor the denom- inator 1 x 2 - 2 x = 1 x ( x - 2) so that the denominator has two linear unrepeated factors, x and ( x - 2). Thus we write 1 x ( x - 2) = A 1 x + A 2 x - 2 , where we need to determine the unknown constants A 1 and A 2 . Multiply both sides of this equality by x ( x - 2) in order to clear the fractions, and we have 1 = A 1 ( x - 2) + A 2 x, (1) or 1 = ( A 1 + A 2 ) x - 2 A 1 , (2) where we have collected together the terms involving x and those involving constants only. But we can write the left-hand side of equation (2) as a ﬁrst degree polynomial in x as well, i.e., 1 = 0 · x + 1, so that (2) becomes 0 · x + 1 = (

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Unformatted text preview: A 1 + A 2 ) x + (-2 A 1 ) . Since the coecients in front of x and the stand-alone constants must equal each other, it follows that 0 = A 1 + A 2 1 =-2 A 1 , a system of two equations in the two unknowns A 1 and A 2 . Solving the second equation for A 1 , we have A 1 =-1 2 . Substituting this value into the rst equation gives 0 =-1 2 + A 2 , or A 2 = 1 2 . Thus, 1 x 2-2 x = 1 x ( x-2) = (-1 2 ) x + ( 1 2 ) x-2 =-1 2 1 x + 1 2 1 x-2 , and Z 1 x 2-x =-1 2 Z dx x + 1 2 Z dx x-2 =-1 2 ln | x | + 1 2 ln | x-2 | + C. 2...
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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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8.3.9-eg - A 1 + A 2 ) x + (-2 A 1 ) . Since the coecients...

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