1. ## Integrating ^-1

I am integrating (1/3)(3-x)^-1.

Is it (1/3)ln(3-x) ?

It probably isn't.

2. Originally Posted by Stuck Man
I am integrating (1/3)(3-x)^-1.

Is it (1/3)ln(3-x) ?

It probably isn't.
$\frac{1}{3}\int \frac{dx}{3-x} dx=-\frac{1}{3} ln|3-x| + C$
You forgot "-" and "C".

3. I'm not sure why you put - at the front.

4/x integrates to 4lnx doesn't it?

4. Originally Posted by Stuck Man
I'm not sure why you put - at the front.

4/x integrates to 4lnx doesn't it?
yes, but $\frac{1}{3-x}$ integrates to $-\ln|3-x| + C$ , because the derivative of $(3-x)$ is $-1$

5. I see, thanks.

Is k sometimes used instead of C?

6. Originally Posted by Stuck Man
I see, thanks.

Is k sometimes used instead of C?
Not usually but it can be-often with logs.

It usually arises when you have the case of $I = ln|x| + C$

because $ln(a)+ln(b) = ln(ab)$ a new constant, k, is introduced such that $C = ln(k)$

$I = ln|x| + C = ln|x| + ln(k) = ln(k|x|)$ and $k|x| = e^I$

7. Originally Posted by Stuck Man
I see, thanks.

Is k sometimes used instead of C?
Any letter can be used to represent a constant- as long as you are clear and consistent.