Your maths looks fine, just don't forget your +C on the final answer
You can simplify using log laws to either:
where
Personally I'd pick the first answer
I need to find the Integral with respect to x of (e^(-.02x))/(8-e^(-.02x))
I use u=8-e^(-.02x)
and find that dx= du/.02e^(-.02x) or dx=50du/e^(-.02x)
when i substitute that back into the integral, the integral reduces to:
integral of (50/u)du which equals 50ln(abs(u))+ c
after substituting back in for u I finished with
50ln(abs(8-e^(-.02x)))
Where did I mess up?
Thanks, I looked back through it and I thought it was ok... Thanks for reminding me about the constant!
Also, I'm new to the forum, I can't copy and past microsoft equation editor into the post, how do you type your equations? Does it automatically format them if you use the proper code?
[tex]CODE[/tex]. Hence my final answer is: [TEX]50\ln\left(|8-e^{-.02x}|\right) + C[/TEX]
The codes for various LaTeX is given inthis thread (although the images don't work so it's either google or the PDF)