Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question sort of follows on from question Functions with logarithmic integrals. The book presents an example of integrating a function whose integral is logarithmic: $$\int \frac{1}{4-3x} dx = -\frac{1}{3}\ln{|4 - 3x|} + K$$

$$= -\frac{1}{3}\ln{A|4 - 3x|}$$

$$= \frac{1}{3}\ln{\frac{A}{|4 - 3x|}}$$

I'm having trouble seeing how the final step is reached. My approach is to separate the logarithm of the product to the addition of separate logs then distribute the minus:

$$-\frac{1}{3}\ln{A|4 - 3x|} = -\frac{1}{3}(\ln{A} + \ln{|4 - 3x|}) = \frac{1}{3}(-\ln{A} - \ln{|4 - 3x|})$$

The I use the property that log minus another log is the log of the first divided by the second to get this:


But I still have a minus that the example in the book doesn't have. Could someone help me with this please? Apologies for asking another question so soon.

share|cite|improve this question
what does $lnA[4-3*x]$ means?can you determine by words?is it product of A and 4-3*x? – dato datuashvili May 6 '12 at 11:32
It could be the case that A in the second step and the final step are not the same. But because it's an integration constant, it is kept as is. So the answer is 1/3*C/|4-3x| where C = 1/A. More like K = + 1/3*ln(A) – zubinmehta May 6 '12 at 11:33
up vote 2 down vote accepted

From $-\dfrac{1}{3}\ln \left( A\left\vert 4-3x\right\vert \right) $ we don't get $\dfrac{1}{3}\ln \frac{A}{\left\vert 4-3x\right\vert }$, because $$\begin{equation*} -\frac{1}{3}\ln \left( A\left\vert 4-3x\right\vert \right) \neq \frac{1}{3} \ln \frac{A}{\left\vert 4-3x\right\vert }. \end{equation*}$$ However if we write the constant of integration $C$ as $C=\frac{1}{3}\ln A$, we get the final result, as follows: $$\begin{eqnarray*} \int \frac{1}{4-3x}dx &=&-\frac{1}{3}\ln \left\vert 4-3x\right\vert +C \\ &=&-\frac{1}{3}\ln \left\vert 4-3x\right\vert +\frac{1}{3}\ln A \\ &=&\frac{1}{3}\left( -\ln \left\vert 4-3x\right\vert +\ln A\right) \\ &=&\frac{1}{3}\ln \frac{A}{\left\vert 4-3x\right\vert }. \end{eqnarray*}$$

share|cite|improve this answer
right because last one is equal to $1/3ln(A)-1/3ln(4-3*x)$ and by using rule of logarithms substraction,great answer @Américo Tavares – dato datuashvili May 6 '12 at 11:46
@dato: Thanks. Yes, by the quotient rule (or subtraction rule). – Américo Tavares May 6 '12 at 11:48
@AmericoTaveres may i ask one question?can i use this site in english language?because i dont know unfortunately spain – dato datuashvili May 6 '12 at 12:01
@dato My blog has only a few posts in English (search for the 'Math' tag). It is mainly in Portuguese. – Américo Tavares May 6 '12 at 12:58
ok thanks @ Américo Tavares – dato datuashvili May 6 '12 at 13:47

generally it would be clear that this one

$$\int \frac{1}{4-3x} dx = -\frac{1}{3}\ln{|4 - 3x|} + K$$ can be transformed into

$-ln[4-3*x]+3*K$ or

i hope this would help you,you should know that in case of $a*ln(f(x))$,$a$ comes in place of power Edited: because i gave answer into different interpretation,let denoted $3*K=ln(A)$,so now by using multiplicatipn rule,$ln(1/(4-3*x))+ln(A)=ln(A/(4-3*x))$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.