Converting Improper Integral with Arctan to Partial Fractions

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SUMMARY

The discussion focuses on solving the improper integral from 1 to infinity of the function (arctan(x)/x^2)dx using integration by parts and partial fractions. The user successfully applies integration by parts with u = arctan(x) and dv = x^-2dx, leading to a partial solution. The challenge arises in transforming the integrand from (1/(x)(1+x^2)) to the form (1/x) - (x/(1+x^2)), which is essential for completing the solution using partial fractions. The discussion confirms the necessity of understanding partial fractions for this integral.

PREREQUISITES
  • Integration by parts
  • Partial fractions decomposition
  • Improper integrals
  • Basic calculus concepts
NEXT STEPS
  • Study the method of integration by parts in detail
  • Learn about partial fraction decomposition techniques
  • Explore improper integrals and their convergence criteria
  • Practice solving integrals involving arctangent functions
USEFUL FOR

Students and educators in calculus, particularly those focusing on integration techniques and improper integrals, as well as anyone seeking to deepen their understanding of partial fractions in mathematical analysis.

Shannabel
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Homework Statement


find the integral from 1 to infinity of (arctanx/x^2)dx


Homework Equations





The Attempt at a Solution


i used integration by parts:
u=arctanx
du=1/(1+x^2)dx
dv=x^-2dx
u=(-1/x)

-arctanx/x + [(1/(x)(1+x^2))dx]from 1 to infinity
i have a partial solution in my book, and here it suggests that i change the integrand to
(1/x)-(x/(1+x^2)) which if i work backward, i can see is equal to the original integrand, but i don't see how to get from (1/(x)(1+x^2)) to (1/x)-(x/(1+x^2))
help?
 
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It's done with partial fractions; have you covered partial fractions before?
 
Bohrok said:
It's done with partial fractions; have you covered partial fractions before?

yes! thankyou :)
 

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