How should i solve the integral of ln(2x-1)/x^2 ?

1 Answer
Jan 10, 2018

#-ln(2x-1)/x+2*(ln(2x-1)-ln(x))+C#

Explanation:

Use integration by parts #intudv=uv-intvdu#

Let #u=ln(2x-1)# and #dv=1/x^2dx#

Then #du=2*1/(2x-1)dx# and #v=-1/x#

#intln(2x-1)*1/x^2dx#

#ln(2x-1)*(-1/x)-int(-1/x)2*1/(2x-1)dx#

#-ln(2x-1)/x+2*int1/(x(2x-1))dx#

Use partial fraction

#-ln(2x-1)/x+2*int2/(2x-1)-1/xdx#

#-ln(2x-1)/x+4*int1/(2x-1)dx-2*int1/xdx#

#-ln(2x-1)/x+2*ln(2x-1)-2*ln(x)+C#