What is the antiderivative of #ln(1+x)#? Calculus Techniques of Integration Integration by Parts 1 Answer Konstantinos Michailidis Mar 10, 2016 The antidervative is #x*ln(1+x)-x+ln(1+x)+c# Explanation: The antiderivative of #ln(1+x)# is the integral #int ln(1+x)dx=int x' *ln(1+x)dx=x*ln(1+x)-int x/(1+x)dx= x*ln(1+x)-int[1-1/[1+x]]dx= x*ln(1+x)-x+ln(1+x)+c# Answer link Related questions How do I find the integral #int(x*ln(x))dx# ? How do I find the integral #int(cos(x)/e^x)dx# ? How do I find the integral #int(x*cos(5x))dx# ? How do I find the integral #int(x*e^-x)dx# ? How do I find the integral #int(x^2*sin(pix))dx# ? How do I find the integral #intln(2x+1)dx# ? How do I find the integral #intsin^-1(x)dx# ? How do I find the integral #intarctan(4x)dx# ? How do I find the integral #intx^5*ln(x)dx# ? How do I find the integral #intx*2^xdx# ? See all questions in Integration by Parts Impact of this question 1480 views around the world You can reuse this answer Creative Commons License