Partial Fractions, Irreducible quadratic factors

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SUMMARY

The discussion focuses on solving the integral for arc length using partial fractions and irreducible quadratic factors. The integral presented is L = ∫(v² dv)/(v² - 1) from √2 to √(1 + e²). The method involves performing long division on the integrand and applying u-substitution, followed by partial fraction decomposition to simplify the expression. Participants suggest using hyperbolic substitution as a potentially quicker alternative to the standard method.

PREREQUISITES
  • Understanding of integral calculus, specifically arc length calculations
  • Familiarity with partial fraction decomposition techniques
  • Knowledge of u-substitution in integration
  • Basic concepts of hyperbolic functions and substitutions
NEXT STEPS
  • Study the method of long division in integrals, particularly for rational functions
  • Learn about partial fraction decomposition with irreducible quadratic factors
  • Explore hyperbolic substitutions in integral calculus
  • Practice solving arc length problems using various integration techniques
USEFUL FOR

Students and educators in calculus, particularly those focusing on integration techniques, as well as mathematicians interested in advanced methods for solving integrals involving rational functions.

rocomath
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Arc Length, Irreducible quadratic factors

i'm having a hard time seeing this method, and i have to use this method on one of the problems I'm doing to find it's Arc Length.

[tex]L=\int_{\sqrt{2}}^{\sqrt{1+e^{2}}}\frac{v^{2}dv}{v^{2}-1}}[/tex]

the book suggests to first divide then use a u-substitution. i know that when the power in the numerator is greater than the denominator, i can perform long-division, but i don't see how i can divide this.
 
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i don't know whether i am gettin u well, but you could do this inegral by doing this to the function under the integral sign

v^2/(v^2-) =(v^2-1+1)/(v^2-1) = (1- 1/(v-1)(v+1) )

then you could use partial fraction by letting

1/(v-1)(v+1) = A/(v-1) + B /(v+1), and then finding the values of B and A. Pardone me if i misunderstood u.
 
Well, like sutupidmath did, add one and then subtract one to the numerator. Then, instead of factoring the new denominator and partial fractions-ing, Try a hyperbolic substitution, I think you will find it quicker.
 

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