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[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 I am 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??

Please help me out! Thanks!

 

[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.

 

[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.

 

[eric_999]

Thanks a lot, i understand now!