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# Can someone help me with this integral?

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1. I've got this far but I can't seem to find the answer.
2. Convert -1/3 / 2x-6 into 1 / 6(x-3).

Then take out a factor of 1/6

Then integrate.
3. HINT:Think about Ln function, rule is when its 1/X or 1/(something) to the power of 1 when you integrate it will be Ln(something) and then the reverse chain rule to get back to it if you were to differentiate
4. Remember
$\int \frac{1}{f(x)} \ dx = \frac{ln\left | f(x) \right |}{f'(x)} + C$
5. (Original post by The_Big_E)
Remember
$\int \frac{1}{f(x)} \ dx = \frac{ln\left | f(x) \right |}{f'(x)} + C$
This is a common misconception among students - the formula you have given applies only when f(x) is linear, i.e. of the form ax+b. For example, let f(x)=e^(x^2), giving the integral of e^(-x^2) - your formula gives some function as the answer, but in fact e^(-x^2) is impossible to integrate, though this result is beyond A-Level.
6. Thanks for all of the help, it took a while but I managed to do It.

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