How do you find the Integral of #ln(2x+1)#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Base e 1 Answer Konstantinos Michailidis Sep 11, 2015 It is #int(ln(2x+1))dx=xln(2x+1)-x+1/2ln(2x+1)# Explanation: We will use integration by parts #intln(2x+1)dx=int(x)'ln(2x+1)dx=xln(2x+1)-intx*(ln(2x+1))'dx=xln(2x+1)-int(x*2/(2x+1))=xln(2x+1)-int(2x+1)/(2x+1)-1/(2x+1)dx=xln(2x+1)-x+1/2ln(2x+1)# Answer link Related questions What is the derivative of #y=3x^2e^(5x)# ? What is the derivative of #y=e^(3-2x)# ? What is the derivative of #f(theta)=e^(sin2theta)# ? What is the derivative of #f(x)=(e^(1/x))/x^2# ? What is the derivative of #f(x)=e^(pix)*cos(6x)# ? What is the derivative of #f(x)=x^4*e^sqrt(x)# ? What is the derivative of #f(x)=e^(-6x)+e# ? How do you find the derivative of #y=e^x#? How do you find the derivative of #y=e^(1/x)#? How do you find the derivative of #y=e^(2x)#? See all questions in Differentiating Exponential Functions with Base e Impact of this question 1671 views around the world You can reuse this answer Creative Commons License