Why Does 1/3 ln((x+1/3)) Seem Incorrect for the Integral of 1/(3x+1)?

[alingy1]

http://www.wolframalpha.com/input/?i=integral+of+1/(3x+1)=1/3ln(x+1/3)

Why is 1/3 ln((x+1/3)) not right?

1/(3x+1)=(1/3)(1/(x+1/3))

Use substitution u=x+1/3, du=dx.

Why is this wrong?

 

[Dick]

http://www.wolframalpha.com/input/?i=integral+of+1/(3x+1)=1/3ln(x+1/3)

Why is 1/3 ln((x+1/3)) not right?

1/(3x+1)=(1/3)(1/(x+1/3))

Use substitution u=x+1/3, du=dx.

Why is this wrong?

$\frac{1}{3 ln(x+1/3)}$ 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!

 

[alingy1]

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

 

[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.

 

[alingy1]

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

 

[Simon Bridge]

Don't trust the machines.

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

 

[alingy1]

Exactly, both are two acceptable results!