How do you integrate #intln(2x+1)dx#? Calculus Techniques of Integration Integration by Parts 1 Answer GiĆ³ May 7, 2015 I would set #2x+1=t# #2dx=dt# #intln(t)dt/2=# by parts: #=1/2[tln(t)-intt*1/tdt]=# #=1/2[tln(t)-t+c]# but #t=2x+1# #=1/2[(2x+1)ln|2x+1|-(2x+1)+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 1350 views around the world You can reuse this answer Creative Commons License