How do you evaluate the indefinite integral #int (-4x^4+x^2-6)dx#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Narad T. Feb 4, 2017 The answer is #=-4/5x^5+1/3x^3-6x+C# Explanation: We use #intx^ndx=x^(n+1)/(n+1)+C(x!=-1)# Therefore, #int(-4x^4+x^2-6)dx# #=-4intx^4dx+intx^2dx-6intdx# #=-4/5x^5+1/3x^3-6x+C# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 2130 views around the world You can reuse this answer Creative Commons License