Integrating Rational Functions with Substitution: How to Solve Tricky Integrals

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SUMMARY

The discussion focuses on solving the integral from 1/3 to 1/2 of the function 1/x(x+1)(x-1)(x^2+1) using substitution and partial fraction decomposition. Participants suggest using the substitution u = x^4 - 1, leading to the transformation of the integral into a more manageable form. The final solution involves integrating the expression 1/[(u+1)(u)], which is decomposed into partial fractions, yielding the result of 1/4 ln(3/16). This method effectively simplifies the integration process and provides a clear path to the solution.

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  • Understanding of integral calculus and definite integrals
  • Familiarity with substitution methods in integration
  • Knowledge of partial fraction decomposition techniques
  • Proficiency in manipulating logarithmic functions
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  • Study the method of substitution in integral calculus
  • Learn about partial fraction decomposition in detail
  • Practice integrating rational functions using various techniques
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Students and educators in calculus, mathematicians focusing on integration techniques, and anyone looking to enhance their skills in solving complex integrals.

  • #31
Shannabel said:
nevermind, i wrote it down opposite from you :)

okay so now i have
1/4 * [ln(u)-ln(u+1)] between 1/3 and 1/2, right?

Yep!

EDIT: err, no. The limits are not right. :redface:
 
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  • #32
I like Serena said:
Yep!

EDIT: err, no. The limits are not right. :redface:

what's wrong with the limits?
i went on and got the right answer:

1/4*[1/u-1/(u+1)]between 1/3 and 1/2
=1/4(lnu-ln(u+1))
=1/4(lnu/(u+1))
when x=1/2, u=-15/16
when x=1/3, u=-80/81
= 1/4 [ln((-15/16)/(1/16))-ln((-80/81)/(1/81))]
= 1/4 [ln(-15)-ln(-80)]
=1/4 [ln(15/80)]
= 1/4ln(3/16)

:)
 
  • #33
Shannabel said:
what's wrong with the limits?
i went on and got the right answer:

1/4*[1/u-1/(u+1)]between 1/3 and 1/2
=1/4(lnu-ln(u+1))
=1/4(lnu/(u+1))
when x=1/2, u=-15/16
when x=1/3, u=-80/81

The limits 1/3 and 1/2 would've been wrong if you used them while the integral was in terms of u, but you changed them right here and those are the correct limits; you didn't tell us you changed them in the previous post :wink:
 
  • #34
What Bohrok said! :approve:
 

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