notes Partial Fractions

Thomas' Calculus: Early Transcendentals

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Partial Fractions Irreducible quadratics in the denominator Suppose that in working a problem by partial fractions you encounter a fraction with irreducible quadratic denominator. How do you integrate it? For example, let I = x + 3 x 2 + 4 x + 7 dx . First, complete the square: x 2 + 4 x + 7 = ( x + 2) 2 + 3. Thus I = x + 3 ( x + 2) 2 + 3 dx . WARNING: Do not split this up as I = x ( x + 2) 2 + 3 dx + 3 ( x + 2) 2 + 3 dx . In the first of these two integrals, the numerator x is not a constant multiple of the derivative 2( x + 2) of the denominator, so substitution does not work. Substitute for the quantity in the completion of the square: let y = x +2, so x = y - 2 and dx = dy . Hence I = y + 1 y 2 + 3 dy . Now split it up. I is the sum of y y 2 + 3 dy = 1 2 ln | y 2 + 3 | + C and 1 y 2 + 3 dy = 1 3 arctan y 3 + C . The final answer, expressed in terms of x is 1 2 ln | x 2 + 4 x + 7 | + 1 3 arctan x + 2 3 + C. (The absolute value sign after ln is unnecessary since x 2 + 4 x + 7 is always positive.) Repeated factors This section will not help you work problems; it is intended to answer a question that many people wonder about. Again let us take an example. Consider dx x ( x + 1) 2 . We are told to set up partial fractions as follows:
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