How Do You Integrate Using Logarithms for \(\int \frac{3}{3x-2} \, dx\)?

[kuahji]

Evaluate the integral.

$$\int3/(3x-2) dx$$ from 0 to -1 (top to bottom).

I change the equation to [tex(1/x - 3/2) dx[/tex]
then integrated ln x-3/2x, but ln x at 0 is undefined.

The textbook shows it as becoming ln (3x-2), but I'm not completely understanding how to get to that.

 

[olgranpappy]

substitute y = 3x-2

 

[kuahji]

Thanks, that worked. I'm just a bit rusty I reckon.

 

[HallsofIvy]

Caution: you are going to find it very difficult to do Calculus if you cannot do basic algebra.

$a/(b+c)$ is not equal to (a/b)+ (a/c)!

You had better review your algebra.

 

[kuahji]

Caution: you are going to find it very difficult to do Calculus if you cannot do basic algebra.

$a/(b+c)$ is not equal to (a/b)+ (a/c)!

Hmm, guess you're right. I've always used (a+b)/c=a/c+b/c but guess I just assumed it would work vice versa. Is there any combination it does equal?

 

[asif zaidi]

My 1st help in answering !

1- You have to use change of variables method.
2- Substitute den with a variable and take derivative of this wrt x
3- Using this var the limits will change
4- Once you do 2 and with new limits from 3, the integral will be easy.

Thanks

Asif