How do you integrate #x(x^2+1)^2 dx#? Calculus Introduction to Integration Integrals of Polynomial functions 1 Answer Anthony R. Mar 16, 2018 #intx(x^2+1)^2dx=x^6/6+x^4/2+x^2/2+"C"# Explanation: Expand #(x^2+1)^2# #(x^2+1)^2=x^4+2x^2+1# Distribute the #x# #x(x^4+2x^2+1)=x^5+2x^3+x# Next we integrate each term #intx^5+2x^3+xdx=intx^5dx+int2x^3dx+intxdx# #=x^6/6+2*x^4/4+x^2/2# #=x^6/6+x^4/2+x^2/2+"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 8490 views around the world You can reuse this answer Creative Commons License