How do you integrate #x/(1+x^4)#? Calculus Introduction to Integration Integrals of Rational Functions 1 Answer Antoine May 5, 2015 #int x/(1+x^4)dx =1/2arctan(x^2)# Method #int x/(1+x^4)dx = 1/2int (2xdx)/(1+(x^2)^2)=1/2int (d(x^2))/(1+(x^2)^2) ="I"# let #x^2 =u# #=>"I"= 1/2int (du)/(1+u^2) = 1/2arctanu = 1/2arctan(x^2)# Answer link Related questions How do you integrate #(x+1)/(x^2+2x+1)#? How do you integrate #dx / (2sqrt(x) + 2x#? What is the integration of #1/x#? How do you integrate #(1+x)/(1-x)#? How do you integrate #(2x^3-3x^2+x+1)/(-2x+1)#? How do you find integral of #((secxtanx)/(secx-1))dx#? How do you integrate #(6x^5 -2x^4 + 3x^3 + x^2 - x-2)/x^3#? How do you integrate #((4x^2-1)^2)/x^3dx #? How do you integrate #(x+3) / sqrt(x) dx#? How do you find the integral of #(x^4+x-4) / (x^2+2)#? See all questions in Integrals of Rational Functions Impact of this question 39815 views around the world You can reuse this answer Creative Commons License