Integration By Parts - ln[3x-2]

[hank]

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?

 

[pka]

The antiderivative of ln(u) is u[ln(u)]-u.

 

[hank]

So just sub in 3x - 2 for the u?

 

[galactus]

Use could use parts and let \(\displaystyle u=ln(3x-2); \;\ dv=dx; \;\ du=\frac{3}{3x-2}dx; \;\ v=x\)

Go through the parts. It's good practice. :wink:

 

[soroban]

Hello, hank!

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

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

Did you try anything?


Take a guess . . .

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

Then: \(\displaystyle \:du\:=\:\text{(something)'}\;\;\;\;\underbrace{v \:=\:\int \ln[3x\,-\,2]\,dx}\)
. . . . . . . . . . . . . . . . . . . . . . .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 . . .