Solution since the power of x in the numerator is the

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Solution : Since the power of x in the numerator is the power in the denominator, we do long division first: x 3 x 2 - 2 x + 1 = x + 2 + 3 x - 2 x 2 - 2 x + 1 . Now, 3 x - 2 x 2 - 2 x + 1 = 3 x - 2 ( x - 1) 2 = A x - 1 + B ( x - 1) 2 = A ( x - 1) + B ( x - 1) 2 . Thus Ax = 3 x and - A + B = - 2 , so A = 3 and B = 1. Partial Fractions (6.3 in e-book)
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Example 4, continued Therefore, Z x 3 x 2 - 2 x + 1 dx = Z h x + 2 + 3 x - 1 + 1 ( x - 1) 2 i dx = x 2 2 + 2 x + 3 ln | x - 1 | - 1 x - 1 + C . Partial Fractions (6.3 in e-book)
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Heavyside “Cover up” method Given a quotient of polynomials f ( x ) / g ( x ), factor g . If g ( x ) = ( x - r 1 )( x - r 2 ) · · · ( x - r n ) , all r i terms different , we can use the Heavyside method. (note: g ( x ) = ( x - 1) 2 and g ( x ) = x 2 + 1 do NOT satisfy this) 1 Write f ( x ) / g ( x ) in its (unsolved) partial fraction expansion f ( x ) g ( x ) = A 1 x - r 1 + A 2 x - r 2 + · · · + A n x - r n . 2 To get A 1 , cover up x - r 1 on the left hand side and evaluate f ( x ) / g ( x ) (disregarding x - r 1 !) at x = r 1 . 3 The same procedure applies to each A i . Partial Fractions (6.3 in e-book)
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Example 5 Find Z 2 x + 1 x 2 - 7 x + 12 dx . Solution : x 2 - 7 x + 12 = ( x - 4)( x - 3), so we can use the cover-up method. Write 2 x + 1 ( x - 4)( x - 3) = A x - 4 + B x - 3 . For A , we cover up x - 4 on the left hand side and set A = h 2 x + 1 x - 3 i x =4 = 9 1 = 9 .
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