How do you evaluate the indefinite integral #int (2x^2-4x+3)dx#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Guillaume L. Apr 13, 2018 #int(2x^2-4x+3)dx=2/3x^3 -2x^2+3x+C, C in RR# Explanation: #int(2x^2-4x+3)dx=int2x^2dx-int4xdx+int3dx# #=2intx^2dx-4intxdx+3intdx# Also, #int(x^n)dx=(x^(n+1))/(n+1)# So :#int(2x^2-4x+3)dx=2*x^3/3-4*x^2/2+3x+C=2/3x^3-2x^2+3x+C, C in RR# \0/ here's our answer ! 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 7297 views around the world You can reuse this answer Creative Commons License