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# integral watch

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1. hi, The integral of 1/3x by the rule (1/a)ln(ax + b) would be 1/3ln3x. But if you take out a third at the beginning you would be left with 1/3 times integral of 1/x which is 1/3 lnx.

What am i doing wrong?
2. (Original post by 111davey1)
hi, The integral of 1/3x by the rule (1/a)ln(ax + b) would be 1/3ln3x. But if you take out a third at the beginning you would be left with 1/3 times integral of 1/x which is 1/3 lnx.

What am i doing wrong?
It's fine to do it either way. Just remember that indefinite integrals have a +C. In the two forms you've mentioned, the C values will be different as
3. (Original post by 111davey1)
hi, The integral of 1/3x by the rule (1/a)ln(ax + b) would be 1/3ln3x. But if you take out a third at the beginning you would be left with 1/3 times integral of 1/x which is 1/3 lnx.

What am i doing wrong?
I guess it's safe to assume you mean in which case, what rule do you mean?? Simply saying "" is not stating any 'rule'

The integral of

If you have then this is the same as which you can surely apply by saying and which indeed gets you but remember that you can split this into which then goes to:

for some arbitrary constant D - which is the same as the first approach.

EDIT: Well I guess I've been beaten to it
4. (Original post by RDKGames)
I guess it's safe to assume you mean in which case, what rule do you mean?? Simply saying "" is not stating any 'rule'

The integral of

If you have then this is the same as which you can surely apply by saying and which indeed gets you but remember that you can split this into which then goes to:

for some arbitrary constant D - which is the same as the first approach.

EDIT: Well I guess I've been beaten to it
Thanks very much. Sorry i should have explained the 'rule'.

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