Integration Using Partial Fractions

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The discussion focuses on integrating the function (x^3 - 8x^2 - 1)/((x+3)(x^2-4x+5)) using partial fractions. The user seeks clarification on how to rewrite the integrand after determining the coefficients A, B, and C in the partial fraction decomposition. Specifically, they are confused about the manipulation of the numerator to facilitate a substitution with u = x^2 - 4x + 5. The breakdown reveals that to maintain the numerator while adjusting for the substitution, a constant is subtracted. The conversation emphasizes understanding the algebraic manipulation involved in the integration process.
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Homework Statement



Integrate (x^3 - 8x^2 - 1)/((x+3)(x^2-4x+5))

Homework Equations



This is an integration by partial fractions.

The Attempt at a Solution



http://www.wolframalpha.com/input/?i=integral+%28%28x^3-8x^2-1%29%2F%28%28x%2B3%29%28x^2-4x%2B5%29%29%29dx

I understand everything except where the integrand is rewritten after finding A, B, and C of the partial fraction decomposition. If anyone can help me understand how the integrand is rewritten that would be great. I just cannot make any sense out of it.

Thanks

Edit: If anyone is unfamiliar with WolframAlpha there is a "show steps" button in the top right corner of the problem statement, this is what I am referring to. It is about the 4th step down, rewrite the integrand.
 
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Here is a screenshot of the part I am asking about.
 

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They're starting with this part of the problem:
\int \frac{14 - 41x}{x^2 - 4x + 5}dx

What they're doing is manipulating things that that a substitution of u = x2 - 4x + 5 (hence du = (2x - 4) dx) will work.

To get a numerator of -41/2*x + 82, which equals -(41/2)(2x - 4), they need to keep the numerator unchanged, so they are subtracting 68.

14 - 41x = -41x + 82 - 68 = -41/2(2x - 4) - 68
 
Yep I see it. I noticed the differential so I thought it was a substitution but I just could not get my head around the manipulation. Thank you for breaking it down for me.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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