How do you integrate #int (x^2(x^3 + 1)^3)dx#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Konstantinos Michailidis Oct 11, 2015 If we expand the product #x^2*(x^3+1)^3# we get #x^11+3x^8+3x^5+x^2# hence we have that #int x^2(x^3+1)^3dx=int (x^11+3x^8+3x^5+x^2)dx=x^12/12+3*x^9/9+3*x^6/6+x^3/3=x^12/12+x^9/3+x^6/2+x^3/3+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 5792 views around the world You can reuse this answer Creative Commons License