I wan't too sure where to put this. It is Calculus, but not University level. Any how,

I need to integrate:

1/(3x+7)

Now I'd be fine with it if the power wasn't -1 seeing as I get 0 on the bottom. I think it has to something do with lnx.

Thanks.

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- Mar 8th 2010, 05:46 AMStephenPocoIntegration help
I wan't too sure where to put this. It is Calculus, but not University level. Any how,

I need to integrate:

1/(3x+7)

Now I'd be fine with it if the power wasn't -1 seeing as I get 0 on the bottom. I think it has to something do with lnx.

Thanks. - Mar 8th 2010, 06:04 AMArchie Meade
That's why it doesn't work..

$\displaystyle \int{\frac{1}{x}}dx=ln|x|$

$\displaystyle \int{\frac{1}{u}}du=ln|u|$

Hence $\displaystyle u=3x+7\ \Rightarrow\ \frac{du}{dx}=3\ \Rightarrow\ du=3dx\ \Rightarrow\ dx=\frac{du}{3}$

The integral is now solved using

$\displaystyle \frac{1}{3}\int{\frac{1}{u}}du=\frac{1}{3}ln|u|=\f rac{1}{3}ln|3x+7|$