What is the antiderivative of #ln(x)^2#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Andrea S. Feb 23, 2017 #int (lnx)^2dx = x(ln^2x -2lnx + 2 ) + C # Explanation: Integrate by parts, using #dx# as differential part: #int (lnx)^2dx = xln^2x -int x d(ln^2x) # #int (lnx)^2dx = xln^2x - 2int x lnx/xdx # #int (lnx)^2dx = xln^2x - 2int lnxdx # Integrating by parts again: #int (lnx)^2dx = xln^2x -2 xlnx + 2 int x d(lnx) # #int (lnx)^2dx = xln^2x -2 xlnx + 2 int x dx/x # #int (lnx)^2dx = xln^2x -2 xlnx + 2 int dx # #int (lnx)^2dx = xln^2x -2 xlnx + 2 x + C # #int (lnx)^2dx = x(ln^2x -2lnx + 2 ) + 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 1376 views around the world You can reuse this answer Creative Commons License