Integration by Parts: integral of ln(3x-2)dx

[warwick]

13. Find the integral of ln(3x-2)dx

u = ln(3x-2)
u' = 3/(3x-2)

v = x
v' = 1 dx

integral of ln(3x-2)dx = ln(3x-2)x - integral of [3/(3x-2)][x]dx

 

[galactus]

You're correct so far. Just need to finish.

\(\displaystyle \L\\\int\frac{3x}{3x-2}dx\)

Rewrite as \(\displaystyle \L\\\frac{2}{3x-2}+1\)

Then you can integrate: \(\displaystyle \L\\2\int\frac{1}{3x-2}dx+\int{1}dx\)

You can use a little u subbing. Let \(\displaystyle \L\\u=3x-2, \;\ \frac{du}{3}=dx\)

Does that make it a little easier?.

 

[warwick]

You're correct so far. Just need to finish.

\(\displaystyle \L\\\int\frac{3x}{3x-2}dx\)

Rewrite as \(\displaystyle \L\\\frac{2}{3x-2}+1\)

Then you can integrate: \(\displaystyle \L\\2\int\frac{1}{3x-2}dx+\int{1}dx\)

You can use a little u subbing. Let \(\displaystyle \L\\u=3x-2, \;\ \frac{du}{3}=dx\)

Does that make it a little easier?.

How did you know to do that little addition trick up there?

My u-substitution looks a little different - 2(1/3) integral of (1/u)3dx = (-2/3) ln (3x-2)

if u = 3x-2. Is that valid? I basically integrated (1/u) du to be ln u.

 

[skeeter]

How did you know to do that little addition trick up there?

method 1 ...

\(\displaystyle \L \frac{3x}{3x-2} = \frac{3x - 2 + 2}{3x - 2} = \frac{3x-2}{3x-2} + \frac{2}{3x-2} = 1 + \frac{2}{3x-2}\)

method 2 ...

use long division to divide 3x by 3x-2 ... I'll spare you the details.

 

[warwick]

How did you know to do that little addition trick up there?

method 1 ...

\(\displaystyle \L \frac{3x}{3x-2} = \frac{3x - 2 + 2}{3x - 2} = \frac{3x-2}{3x-2} + \frac{2}{3x-2} = 1 + \frac{2}{3x-2}\)

method 2 ...

use long division to divide 3x by 3x-2 ... I'll spare you the details.

My question wasn't HOW did you do that little addition trick. My question was how did you KNOW to do it. :wink:

 

[skeeter]

when integrating a rational function, if the degree of the numerator > degree of the denominator, perform the long division.

:wink: back at ya'

 

[warwick]

when integrating a rational function, if the degree of the numerator > degree of the denominator, perform the long division.

:wink: back at ya'

Okkk.....