How to find the equation of the tangent at a specific point? PLEASE HELP, 10 POINTS

Find the equation of the tangent to the curve y=ln(3x-2) at the point where x=1. Then find the value of the constant A such that 6x/(3x-2) = 2+ A/(3x-2) and show that the integration with boundaries x=6 and x=2 of 6x/(3x-2) dx = 8+ 8/3 ln2.
Please help me solve this problem, 10 points??

y = ln(3x-2)
Slope = dy/dx = [1/(3x-2)](3) = 3/(3x-2)
When x = 1, slope = 3/[3(1)-2]=3
When x = 1, y = ln[3(1)-2] = 0
Tangent equation:
y = 3x

6x/(3x-2) = 2+ A/(3x-2)
A = {[6x/(3x-2)] – 2}(3x-2)
A = {[6x – 2(3x-2)]/(3x-2)}(3x-2)
A = 6x – 2(3x-2)
A = 6x – 6x + 4
A = 4

Integral 6,2: 6x/(3x-2) dx
= integral 6,2: 2+ 4/(3x-2) dx
= 6,2: {2x + [4ln(3x-2)]/3}
= {2(6) + [4ln(3(6)-2)]/3} - {2(2) + [4ln(3(2)-2)]/3}
= 12 + [4ln(16)]/3 – 4 – [4ln(4)]/3
= 8 + [4ln(2^4)]/3 – [4ln(2^2)]/3
= 8 + [16ln(2)]/3 – [8ln(2)]/3
= 8 + [8ln(2)]/3