# Integral calculation problem

1. Jan 12, 2014

### alingy1

2. Jan 12, 2014

### Dick

$\frac{1}{3 ln(x+1/3)}$ isn't right. $\frac{1}{3} ln(x+1/3)$ is right. Which one are your trying to write anyway? Use parentheses if you don't want to use TeX!

3. Jan 12, 2014

### alingy1

Yes, the latter is the one! But, the computer program says it is wrong! Am I going crazy?

4. Jan 12, 2014

### Simon Bridge

the wolfram alpha input (link) was: integral of 1/(3x+1)=1/3ln(x+1/3)
the output actually tells you why they think this relation is false.

Namely: $\ln\big(3x+1\big)\neq\ln\big(x+\frac{1}{3}\big)$

But it doesn't have to be - equal, that is - since this is an indefinite integral, the two proposed solutions need only be the same to within an arbitrary constant. This is easy to show:

$\ln[x+\frac{1}{3}]=\ln[\frac{1}{3}(3x+1)]=\ln[3x+1]-\ln(3) = \ln[3x+1]+c$

... you have to be careful with indefinite integrals.

5. Jan 12, 2014

### alingy1

Awesome! That's what's been missing! I spent an hour just on this!

6. Jan 12, 2014

### Simon Bridge

Don't trust the machines.

You could have seen their result by using the substitution u=3x+1.

7. Jan 12, 2014

### alingy1

Exactly, both are two acceptable results!