course MTH 272 05/08,1830 assignment #017017.
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21:41:38 `questionNumber 170000 Query problem 6.3.18 integrate 3/(x ^ 2 - 3x)
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RESPONSE --> 3/ (x^2 - 3x) 3/ x(x - 3) 3/x(x-3)= A/x + B/(x-3) Mulitply all by LCD : x(x-3) Basic Equation: 3=A(x-3) + Bx x=0 3=A(0-3) + B(0) 3=A(-3) -1=A x=3 3=A(3-3) + B(3) 3=A(0) + B (3) 1=B int. -1/x + 1/(x-3) -ln|x| + ln|x-3| + C confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:42:38 `questionNumber 170000 What is your integral?
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RESPONSE --> int. -1/x + 1/(x-3) int. -1/x + int. 1/(x-3) confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:46:14 `questionNumber 170000 Explain how you used partial fractions to obtain your result.
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RESPONSE --> By factoring the denominator (x^2 - 3x) into x(x-3), there are now two parts to the denominator. The variables A and B are placed over one of the parts of the denominator. (A/x + B/(x-3)). After multiplying both sides of the equation by the lowest common denominator x(x-3), the basic equation is found. Having found the basic equation, the value of each variable is determined. Placing these values back over their respective part of the denominator, integration can then take place. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:52:50 `questionNumber 170000 Query problem 6.3.29 (was 6.3.27) integrate (x+2) / (x^2 - 4x)
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RESPONSE --> integrate (x+2) / (x^2 - 4x) (x+2)/x(x-4) x+2/ x(x-4)= A/x + B/(x-4) Multiply all by LCD:x(x-4) Basic Equation: x+2= A(x-4) + Bx x=0 0+2=A(0-4) + B(0) 2=A(-4) (-1/2)=A x=4 4+2=A(4-4) + B(4) 6=B(4) (3/2)=B (-1/2)/x + (3/2)/(x-4) (-1/2) int. (1/x) + (3/2) int. 1/(x-4) (-1/2) ln |x| + (3/2) ln |x-4| + C confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:53:36 `questionNumber 170000 What is your integral?
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RESPONSE --> (-1/2) int. (1/x) + (3/2) int. (1/(x-4)) confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:54:48 `questionNumber 170000 Explain how you used partial fractions to obtain your result.
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RESPONSE --> Explanation similar as in previous problem and shown step-by-step in answer. Partial fractions provide a simpler equation from which to begin integration. confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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21:55:02 `questionNumber 170000 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> None. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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