By the laws of logarithms:
#=intlnx^2 + lnsqrt(4x - 1)dx#
#=intlnx^2dx + intlnsqrt(4x - 1)dx#
#=2intlnxdx+ 1/2intln(4x - 1)dx#
We let #u = 4x - 1#. Then #du = 4dx# and #dx = (du)/4#.
#=2intlnxdx + 1/2intlnu (du)/4#
#=2intlnxdx + 1/8intlnudu#
We need to integrate the natural logarithm function by parts. Let #u = lnx# and #dv = 1dx#. Then #du = 1/xdx# and #v = x#.
#int(udv) = uv - int(vdu)#
#int(lnx) = xlnx - int(x * 1/x)dx#
#int(lnx) = xlnx - x + C#
Back to the original integral.
#=2(xlnx - x) + 1/8(u lnu -u)+ C#
#=2xlnx - 2x + 1/8(4x - 1ln(4x - 1) - (4x - 1)) + C#
#=2xlnx - 2x + ((4x - 1)ln(4x - 1) - 4x + 1)/8 + C#
Hopefully this helps!
