Whats wrong with my work for improper integral?

[DeadOriginal]

Homework Statement

∫^{∞}_{1}\frac{dx}{x(2x+5)}

Homework Equations

Mathematica told me that the answer was \frac{1}{5}nl(\frac{7}{2})

The Attempt at a Solution

This is my work.

∫^{∞}_{1}\frac{dx}{x(2x+5)} = lim_{M\rightarrow∞}∫^{M}_{1}\frac{dx}{x(2x+5)} = lim_{M\rightarrow∞}∫^{M}_{1}\frac{\frac{1}{5}}{x}+\frac{\frac{-2}{5}}{2x+5} = lim_{M\rightarrow∞}((\frac{1}{5}∫^{M}_{1}\frac{1}{x})-(\frac{2}{5}∫^{M}_{1}\frac{1}{2x+5})) = lim_{M\rightarrow∞}\frac{1}{5}(nl(x)-nl(2x+5))^{M}_{1} = lim_{M\rightarrow∞}\frac{1}{5}((nl(M)-nl(2M+5))-(nl(1)-nl(7))) = lim_{M\rightarrow∞}\frac{1}{5}((nl(M)-nl(2M+5))-(0-nl(7))) = \frac{1}{5}((∞-∞)+nl(7)) = \frac{1}{5}nl(7) .

I'm looking at it over and over again but I just can't seem to figure out where that nl(2) from mathematica came from. Could someone please look at my work and tell me what is wrong with it? I'd greatly appreciate it.

 

[SammyS]

I've never seen natural log as nl. Usually it's either ln, or simply log.

Where you have nl(1), you should have nl(2) for nl(2x+5) evaluated at x = 1.

That should do it !

 

[DeadOriginal]

I've never seen natural log as nl. Usually it's either ln, or simply log.

Where you have nl(1), you should have nl(2) for nl(2x+5) evaluated at x = 1.

That should do it !

Hi SammyS,

Thanks for the help. I knew nl looked kind of weird. haha. For the sake of keeping it from getting confusing I'll finish this thread with it.

I don't understand what you mean though. For nl(1), I had nl(x) and I evaluated at x = 1. That's why I had nl(1). Shouldn't nl(2x+5) evaluated a x = 1 be nl(7)?

 

[SteamKing]

You have a problem evaluating the upper limit of the integral as M approaches infinity. In your next to last expression, infinity - infinity is not equal to zero.

It would be easier to determine the indefinite integral before applying the limits of integration.

 

[SammyS]

Hi SammyS,

Thanks for the help. I knew nl looked kind of weird. haha. For the sake of keeping it from getting confusing I'll finish this thread with it.

I don't understand what you mean though. For nl(1), I had nl(x) and I evaluated at x = 1. That's why I had nl(1). Shouldn't nl(2x+5) evaluated a x = 1 be nl(7)?

Yes, you are correct. I merely had a brain cramp. (DUH !)

Look at SteamKing's post. regarding ∞ - ∞. I think that's where you may find the ln(2) you're looking for.

 

[DeadOriginal]

OOOOOOOOOOHHHHHHHHHHHHHHHHHHH. *bangs head on table*

Thanks a lot.