Integration By Parts - ln[3x-2]

hank

Junior Member
Joined
Sep 13, 2006
Messages
209
Hi, everyone. New semester, so new questions!

I have this problem where I need to find the integral of ln[3x-2] dx, but I'm not sure which to use for u or which to use for v.

How do I do this one?
 
The antiderivative of ln(u) is u[ln(u)]-u.
 
Use could use parts and let u=ln(3x2);   dv=dx;   du=33x2dx;   v=x\displaystyle u=ln(3x-2); \;\ dv=dx; \;\ du=\frac{3}{3x-2}dx; \;\ v=x

Go through the parts. It's good practice. :wink:
 
Hello, hank!

Find: \(\displaystyle \L\:\int\)ln[3x2]dx\displaystyle \ln[3x\,-\,2]\, dx

but I'm not sure which to use for u\displaystyle u or which to use for v\displaystyle v. . Really?

Did you try anything?


Take a guess . . .

Let u=something\displaystyle u \:=\:\text{something} . . . . . . . dv=ln[3x2]dx\displaystyle dv\:=\:\ln[3x\,-\,2]\,dx

Then: du=(something)’        v=ln[3x2]dx\displaystyle \:du\:=\:\text{(something)'}\;\;\;\;\underbrace{v \:=\:\int \ln[3x\,-\,2]\,dx}
. . . . . . . . . . . . . . . . . . . . . . .
We can't integrate this!

. . Of course not . . . That's the original problem, isn't it?



We'd better try it the other way:

\(\displaystyle \:\begin{array}{cc}u \:=\:\ln[3x\,-\,2]\;\; & \;\;dv\:=\:dx \\
du\:=\:\frac{3}{3x\,-\,2}\,dx\;\; & \;\; v \:=\:x \end{array}\)

See? .It was not a hard decision . . .

 
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