# Thread: Evaluate the Indefinite Integral

1. ## Evaluate the Indefinite Integral

Need some help with this indefinite integral:

The answer I got was d/m (ln(mx+k)) by using u = mx+k.

2. Originally Posted by statelyplump
Need some help with this indefinite integral:

The answer I got was d/m (ln(mx+k)) by using u = mx+k.
d/m (ln(mx+k)) ???

It should be:

$\frac{1}{m} \times ln(mx+k) = \frac{ln(mx+k)}{m}$

3. Originally Posted by harish21
d/m (ln(mx+k)) ???

It should be:

$\frac{1}{m} \times ln(mx+k) = \frac{ln(mx+k)}{m}$
So is d not to be treated as a constant? I'm confused by the notation.

4. Originally Posted by statelyplump
So is d not to be treated as a constant? I'm confused by the notation.
d is not a constant. "dx" means that you are integrating the function with respect to x!

its..

$\int \frac{1}{mx+k} dx$

so when you substituted $u = mx+k$, you should have gotten:
$du = m dx \implies dx = \frac{1}{m} du$

then you substitute $dx = \frac{1}{m} du$ in your question to get:

$\frac{1}{m} \int \frac{1}{u} du$

do you get it?

5. Thanks. I understand everything else, I just didn't know that they were putting dx in the numerator like that. In every other problem, the dx is off to the side like it was when you reformatted it.

6. Originally Posted by statelyplump
Thanks. I understand everything else, I just didn't know that they were putting dx in the numerator like that. In every other problem, the dx is off to the side like it was when you reformatted it.
$\frac{dx}{x}$ and $\frac{1}{x} dx$ are the same!