Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given the very simple integral

\begin{equation} \int -\frac{1}{2x} dx \end{equation}

The obvious solution is

\begin{equation} \int -\frac{1}{2x} dx = -\frac{1}{2} \int \frac{1}{x} dx = -\frac{1}{2} \ln{|x|} + C \end{equation}

However, by the following integration rule \begin{equation} \int \frac{1}{ax + b} dx = \frac{1}{a} \ln{|ax + b|} + C \end{equation}

the following solution is obtained \begin{equation} \int -\frac{1}{2x} dx = -\frac{1}{2}\ln{|-2x|} + C \end{equation}

Why are these solutions different? Which is correct?

The second solution can be simplified \begin{equation} -\frac{1}{2}\ln{|-2x|} + C = -\frac{1}{2}\ln{|-2|} -\frac{1}{2}\ln{|x|} + C= -\ln{\frac{1}{\sqrt{2}}} - \frac{1}{2}\ln{|x|} + C \end{equation} but they still differ.

share|improve this question
Hint: $\ln(xy)=\ln x+ \ln y$ –  tetori Jan 18 '13 at 17:52
$|a|=|-a|$ for any $a\in C$ –  lab bhattacharjee Jan 18 '13 at 17:53

3 Answers 3

They are both correct: they differ only in terms of their constants of integration.

$$\begin{align} -\frac{1}{2}\ln|-2x| + \color{red}{C} &= -\frac{1}{2}(\ln 2 + \ln |x|) + \color{red}{ C} \\&= -\frac{1}{2}\ln |x| + \color{red}{(C - \frac{1}{2} \ln 2)}\\ &= -\frac{1}{2}\ln |x| + \color{red}{K} \end{align} $$

$C \neq K\;$ but $\;C, K\;$ are constants nonetheless!

TIP: One can always check two apparently different solutions to an integral by differentiating each of them; if the respective derivatives are equal to the original integrand, then you can conclude that the two apparently different solutions are, in fact, solutions that differ only in their constants of integration.

EDIT: You were almost there (in obtaining the adjusted constant of integration), but : $$-\frac{1}{2}\ln{|-2|} + C\; \ne \;-\ln{\frac{1}{\sqrt{2}}} + C$$ Rather, since $\;|-2| \;= \;2,\;$ we have $$-\frac{1}{2}\ln{|-2|}+C \; =\; -\frac{1}{2}\ln 2 + C\;=\; +\ln{\frac{1}{\sqrt{2}}} + C = K$$

The important thing to note is that all of $K$ is a constant term.

share|improve this answer
Glad to see you here online , amWhy. ;-) –  B. S. Jan 18 '13 at 18:10
Wooow! You made it colored. ;-) –  B. S. Jan 19 '13 at 17:18

They are both correct!

$$ -\frac{1}{2}\ln|-2x| + C = -\frac{1}{2}(\ln 2 + \ln |x|) + C = -\frac{1}{2}\ln |x| + (C - \frac{1}{2} \ln 2) $$

The constant of integration is what "differs" here.

share|improve this answer

Both of the results is Ok. Note that there is no need that two constants $C$ in the first result and $C$ in the second one are the same. Here, we have $C_1=C_2\times0.5\ln|2|$.

share|improve this answer
+1 Nice to see you too, BabaK! –  amWhy Jan 18 '13 at 18:12

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.