What is the improper integral (arctan2x)/(pi(1+4x^2)) dx from 0 to infinity ?

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Answer 1 · 2018-06-04T21:55:48

#int_0^oo (arctan(2x))/(pi(1+4x^2))dx = pi/16#

Substitute #t=2x#:

#int_0^oo (arctan(2x))/(pi(1+4x^2))dx = 1/(2pi) int_0^oo (arctant)/(1+t^2)dt#

Substitute now:

#u = arctan t#

#du = dt/(1+t^2)#

#int_0^oo (arctan(2x))/(pi(1+4x^2))dx = 1/(2pi) int_0^(pi/2)u du#

#int_0^oo (arctan(2x))/(pi(1+4x^2))dx = 1/(4pi) [u^2]_0^(pi/2)#

#int_0^oo (arctan(2x))/(pi(1+4x^2))dx = pi/16#