

Subject:
calculus integrations
Category: Science > Math Asked by: kdmacga List Price: $10.00 
Posted:
08 Jun 2006 17:27 PDT
Expires: 08 Jul 2006 17:27 PDT Question ID: 736544 
Integral calculus. Solve with explanations 2 integrations. 1. "x^2+1/x^2" and "x^31/x1". The method is the important part. 

Subject:
Re: calculus integrations
Answered By: liviofloresga on 08 Jun 2006 21:14 PDT 
Hi!! Since the commenter suggests, the best method to find the antiderivatives of "x^2+1/x^2" and "x^31/x1" is by simplifying the rational expressions and then apply the basic rules of integration. 1. "x^2+1/x^2" (x^2+1)/x^2 = 1 + 1/x^2, then: (int means integral) int(x^2+1)/x^2 dx = int (1 + 1/x^2) dx = = int dx + int 1/x^2 dx = = (x+C) + (1/x+D) = = x  1/x + E where C, D, E are unknown constants and E=C+D . 2. "x^31/x1" Note that x=1 is a root of (x^31), the (x1) is a divisor of (x^31): x^3  1 = (x1)*(x^2 + x + 1) Then: (x^3  1) / (x1) = (x^2 + x + 1) , so you have that: int (x^31/x1)dx = int (x^2 + x + 1)dx = = int x^2 dx + int x dx + int dx = = x^3/3 + x^2/2 + x + K where K is an unknown constant. The following pages will help you to better understand how to simplify rational expressions: "Simplify Rational Expressions": http://www.analyzemath.com/Rational_expressions/Simp_rat_expre.html "College Algebra Tutorial on Simplifying Rational Expressions": http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut8_simrat.htm "Rational Functions": http://daphne.palomar.edu/mmumford/5660notesPDF/chap8.pdf Search strategy: My own knowledge when solving the problems adn then use the following keywords at Google.com: simplifying rational expressions I hope this helps you. feel free to request for a clarification if you need it. Regards, liviofloresga 

Subject:
Re: calculus integrations
From: kingalga on 08 Jun 2006 20:15 PDT 
1) I assume you mean (x^2 + 1)/x^2. Simplify to 1 + 1/x^2, then integrate. 2) Divide x^3 1 by x1, then integrate that. 
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