How do you find the indefinite integral of #int (-4/x^3-8/x^5) dx#? Calculus Introduction to Integration Integrals of Polynomial functions 1 Answer Narad T. Nov 27, 2016 The answer is #=2/x^2+2/x^4+C# Explanation: We use #intx^ndx=x^(n+1)/(n+1)+C (n!=-1)# So, #int(-4/x^3-8/x^5)dx=int(-4x^(-3)-8x^(-5))dx# #=-4*x^(-3+1)/(-3+1)-8*x^(-5+1)/(-5+1)+C# #=2/x^2+2/x^4+C# Answer link Related questions How do you evaluate the integral #intx^3+4x^2+5 dx#? How do you evaluate the integral #int(1+x)^2 dx#? How do you evaluate the integral #int8x+3 dx#? How do you evaluate the integral #intx^10-6x^5+2x^3 dx#? What is the integral of a constant? What is the antiderivative of the distance function? What is the integral of #|x|#? What is the integral of #3x#? What is the integral of #4x^3#? What is the integral of #sqrt(1-x^2)#? See all questions in Integrals of Polynomial functions Impact of this question 2422 views around the world You can reuse this answer Creative Commons License