# Integration problems

1. Jul 24, 2014

### eric_999

Hey!

I have this integral: ∫((1/2)/(2x-1))dx.

The first time, I did like this: ∫((1/2)/(2x-1))dx = (1/2)∫(1/(2x-1))dx. If I set u = 2x-1, then du = 2dx, so I can rewrite (1/2)∫(1/(2x-1))dx as (1/2)*(1/2)∫(1/u)du = 1/4∫(1/u)du = 1/4ln|u| = 1/4ln|2x-1|.

But when I do like this (I cannot see what im doing wrong) ∫((1/2)/(2x-1))dx = ∫(1/(4x-2)dx and set u = 4x-2, so du = 4dx, I can rewrite it as 1/4∫(1/u)du = 1/4ln|u| = 1/4ln|4x-2|. What am I doing wrong??

2. Jul 24, 2014

### da_nang

The error you made is that you forgot the constant of integration. Include that, and recall that $\ln{ab} = \ln{a} + \ln{b}$ for positive, non-zero a and b, you should be able to see that the methods are equivalent.

3. Jul 24, 2014

### jbunniii

Note that
\begin{align} \frac{1}{4}\ln|4x-2| &= \frac{1}{4}\ln(2|2x-1|) \\ &= \frac{1}{4}(\ln(2) + \ln|2x-1|)\\ &= \frac{1}{4}\ln|2x-1| + \frac{\ln(2)}{4} \end{align}
which differs from your other answer $\frac{1}{4}|2x-1|$ by a constant. Since indefinite integrals can differ by a constant ($+C$), both answers are equally valid.

4. Jul 24, 2014

### eric_999

Thanks a lot, i understand now!