#int (4x)/(1-x^4)dx# =? Calculus Introduction to Integration Integrals of Rational Functions 1 Answer Konstantinos Michailidis Jun 1, 2016 Notice that #4(x/(1-x^4))=2x/( (x^2+1))-1/( (x-1))-1/( (x+1))# Hence #int x/(1-x^4)dx=2int x/( (x^2+1))dx-int1/( (x-1))dx-int1/( (x+1))dx=log(x^2+1)-log(x-1)-log(x+1)+c= log(x^2+1)-log(x^2-1)+c# Answer link Related questions How do you integrate #(x+1)/(x^2+2x+1)#? How do you integrate #x/(1+x^4)#? 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#? See all questions in Integrals of Rational Functions Impact of this question 2510 views around the world You can reuse this answer Creative Commons License