Question

I'm having trouble understanding the expansion of a logarithm and why the argument of the denominator of the original log, now as a standalone term, becomes positive. Posted below in just a minute.

Answer 1

\[2(-\frac{ 1 }{ 2 } \ln( \frac{ 1+(2x+1) }{ 1-(2x+1) }) = -\ln \frac{ (2x+2) }{ (-2x) } = -(\ln(x+1) - \ln(-x) = \ln(-x) - \ln(x+1)\]For whatever reason, the second to last term, ln(-x) should have a positive argument. Wolfram|Alpha says this, along with my book and a few other things. I don't understand why this is the case. Could somebody help me understand this?

Answer 2

God have mercy when I reach this stuff D:>

Answer 3

Hmm weird. Can you write what the exact answer would be? Maybe looking at that could help me explain it (the fact that the problem is cut off for some reason really bugs me too)

Answer 4

lmao, this is coming from the inverse hyperbolic tangent.
\[\tanh ^{-1}(x) = \frac{ 1 }{ 2 }\ln (\frac{ 1+x }{ 1-x })\]

Answer 5

Okay, yeah, one minute for the exact answer.

Answer 6

\[\ln(x)-\ln(x+1)\]

Answer 7

The (sort of, this is a really long problem) "original" thing this stems from is an expansion of \[\tanh ^{-1}(2x+1)\]

Answer 8

Yeah thats very weird. The only thing that I can tell you is that Natural longs cant really have an argument that is negative; it creates an invalid answer/output

Answer 9

Sorry; I really wasnt much help

Answer 10

No worries.

Answer 11

This stems from an integral (btw the original problem just is multiplied by two, the constant doesn't affect the outcome other than everything being twice the value), if we want to start completely from the ground up. I understand everything they do up until this point. They're seeking the integral of \[\int\limits_{}^{}\frac{ 2 }{ x^2 + x }dx\]
First, they start off by factoring out the constant:\[2\int\limits_{}^{}\frac{ 1 }{ x ^{2}+x }\]
Then they complete the square of the denominator:\[2\int\limits_{}^{}\frac{ 1 }{ (x+\frac{ 1 }{ 2 })^{2}-\frac{ 1 }{ 4 } }\]
Then, they use u-substitution:\[u = (x+\frac{ 1 }{ 2 }), du = dx\]\[2\int\limits_{}^{}\frac{ 1 }{u ^{2}+\frac{ 1 }{ 4 } }\]

Answer 12

(One second, cont'd)

Answer 13

Then they factor out the constant 1/4 in the denominator:
\[2\int\limits_{}^{}\frac{ 4 }{ u ^{2}-1 }\]
Then they factor out the constant 4 in the numerator:
\[8\int\limits_{}^{}\frac{ 1 }{ 4u^{2}-1 }\]Then they perform a second u-substitution after factoring out (-1) from the denominator and factoring back in 2 to accomodate this second substitution:
\[s = 2u, ds = 2du\]
\[-4\int\limits_{}^{}\frac{ 1 }{ 1-s ^{2} }ds\]

Then they integrate that into an inverse hyperbolic tangent function and substitute backwards twice in with u and eventually x, and end up with
\[-4 \tanh ^{-1}(2x+1)\]

Could somebody *PLEASE* help me understand what's wrong with this last part when expanding the inverse hyperbolic tangent function?

Answer 14

@agent0smith @ganeshie8 @jim_thompson5910 @Psymon @Mertsj

Answer 15

(Also,sorry for all the missing d[some variable]'s, thought it would be implied)

Answer 16

Don't try to post long equations all at once\[\Large 2(-\frac{ 1 }{ 2 } \ln( \frac{ 1+(2x+1) }{ 1-(2x+1) }) = -\ln \frac{ (2x+2) }{ (-2x) } =\]
\[\Large -(\ln(x+1) - \ln(-x) = \ln(-x) - \ln(x+1)\]

Answer 17

Yeah, that's it. Sorry about that. But treating that as just a standalone problem, the argument of the second to last logarithm should be positive and I don't know why.

Answer 18

(There's also a missing end parentheses in the final line, that closes together both logs inside the negative sign.)

Answer 19

I used a W|A widget, along with the site, to get this answer. Here's the page that shows the simplification of the original log, except with the negative on the outside of the two, which, AFAIK, shouldn't change anything. http://www.wolframalpha.com/input/?i=simplify+-2%281%2F2+ln%28%281%2B2x%2B1%29%2F%281-%282x%2B1%29%29%29%29

Answer 20

(also, I'm ignoring complex numbers, we don't deal with them in this class)

Answer 21

http://www.wolframalpha.com/input/?i=2%28-%5Cfrac%7B+1+%7D%7B+2+%7D+%5Cln%28+%5Cfrac%7B+1%2B%282x%2B1%29+%7D%7B+1-%282x%2B1%29+%7D%29

they don't simplify it near to what you do...

Answer 22

Why does mine show that they do? (second answer down, with an imaginary number and pi in the answer) http://www.wolframalpha.com/input/?i=simplify+-2%281%2F2+ln%28%281%2B2x%2B1%29%2F%281-%282x%2B1%29%29%29%29

Answer 23

still not sure what problem you're seeing with the simplifying. One of them shows an imaginary number which comes from ln(-1).

Answer 24

...why does ln(-x) become ln(x).

Answer 25

I thought I had said that pretty clearly, I guess not.

Answer 26

Because ln(ab) = lna + lnb

ln(-x) = ln(-1) + lnx

Answer 27

Okay. The other widget just didn't have the imaginary number at all, if I run into an answer with some (but not all) imaginary terms, how should I deal with that?

Answer 28

The widget made a mistake then... you can't just simplify it and ignore the ln(-1).

I don't know how you'd deal with them if the class isn't. Just don't simplify to the point where you'd have to introduce them, i guess.

Answer 29

Okay, cool. Thanks!