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.

When you calculate $\int\frac{1}{2x}dx$ you get $\frac{1}{2}\ln(x)$ and when you calculate $\int\frac{1}{2x}dx$ you don't get $\frac{1}{2}\ln(2x)$.

$\frac{1}{2x}$ is the same as $\frac{1}{2x}$

why do you get different answers?

share|improve this question
1  
" $\frac{1}{2x}$ is the same as $\frac{1}{2x}$ ". And $ \frac{1}{2} \ln (x) \ $ is the same as $ \frac{1}{2} \ln (2x) \ , $ with a "vertical shift" when you graph the two functions. When we calculate indefinite integrals, we are not finding a single function, but an entire "family" of functions which differ only by a constant; so, they all have the same derivative function. That is the meaning of the arbitrary constant, as the posters are describing in their answers. –  RecklessReckoner Mar 28 at 13:18

2 Answers 2

$$\frac{1}{2}\ln(2x) = \frac{1}{2}(\ln 2 + \ln x) = \frac{\ln2}{2} + \frac12\ln x$$

When you integrate $$\frac{\frac12}{x},$$ you get $\frac12 \ln x + C$ ($C$ is the constant you are always nagged about!), not just $\frac12\ln x$

share|improve this answer
2  
+1. Never forget the constant of integration when doing indefinite integrals. –  Arthur Mar 28 at 12:59
2  
Questions like this are very good in my oppinion, as they really show that the $+C$ is not just for show. –  5xum Mar 28 at 12:59
    
(+1) Good answer. –  Thomas Mar 28 at 13:11

I think that @5xum has already said it, but to make it super clear:

The integral of $1 / (2x)$ is not equal to $\frac{1}{2}\ln(x)$ or $\frac{1}{2}\ln(2x)$. you have $$ \int \frac{1}{2x} \; dx = \frac{1}{2}\ln(x) + C. $$ As noted in the other answer and in the comments, this $+C$ isn't just a pretty thing. It is very important because without it, the answer is wrong. The example that you have given shows exactly why it is important. So you have, for example, $$ \int \frac{1}{2x} \; dx = \frac{1}{2}\ln(x) + C_1 = \frac{1}{2}\ln(2x) + C_2 $$ where $C_1$ and $C_2$ are different constants. The reason that this is true is because of the definition. Remember that we say that $$ \int f(x)\; dx = F(x) + C $$ exactly when $F'(x) = f(x)$. And indeed you have that $$\begin{align*} \frac{d}{dx} \frac{1}{2}\ln(x) = \frac{d}{dx} \frac{1}{2}\ln(2x) = \frac{1}{2x}. \end{align*} $$

share|improve this answer
    
It's not so much that the answer is "wrong", as that the result is incomplete. An indefinite integral produces an infinite set of anti-derivative functions, all having the same derivative function, which is the integrand. –  RecklessReckoner Mar 28 at 13:24
    
already accepted answer but this is great :) +1 –  Yusaf Mar 28 at 13:34

Your Answer

 
discard

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.