Homework Help: 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!